Anti-Derivatives Exercise: 19.2 Class 11 Basic Mathematics Solution [NEB UPDATED]

Exercise 19.2

Integrate the following:

1.(i) $\mathop \smallint \nolimits 3{{\rm{x}}^2}{\left( {{{\rm{x}}^3} + 1} \right)^3}$.dx

Solution:

Let I = $\mathop \smallint \nolimits 3{{\rm{x}}^2}{\left( {{{\rm{x}}^3} + 1} \right)^3}$.dx

Put x³ + 1 = y then 3x².dx = dy

Or, I = $\mathop \smallint \nolimits {{\rm{y}}^3}$.dy = $\frac{{{{\rm{y}}^4}}}{4}$ + c = $\frac{1}{4}$ (x³ + 1)⁴ + c

 

(ii) = $\mathop \smallint \nolimits {\rm{x}}{\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)^{\frac{3}{2}}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\rm{x}}{\left( {{{\rm{a}}^2} + {{\rm{x}}^2}} \right)^{\frac{3}{2}}}$.dx

Put a² + x² = y then 2x.dx = dy

Or, x.dx = $\frac{1}{2}$.dy

Or, I = $\mathop \smallint \nolimits {{\rm{y}}^{\frac{3}{2}}}.\frac{1}{2}$.dy

= $\frac{1}{2}\mathop \smallint \nolimits {{\rm{y}}^{\frac{3}{2}}}$.dy = $\frac{1}{2}\frac{{{{\rm{y}}^{\frac{5}{2}}}}}{{\frac{5}{2}}}$ + c

= $\frac{1}{5}$y^5/2 + c = $\frac{1}{5}$ (a² + x²)^5/2 + c.

 

(iii) $\mathop \smallint \nolimits \frac{{{\rm{xdy}}}}{{3{{\rm{x}}^2} - 4}}$.dx
Solution:

Let I = $\mathop \smallint \nolimits \frac{{{\rm{xdy}}}}{{3{{\rm{x}}^2} - 4}}$.dx

Put 3x² – 4 = y

Or,6x.dx = dy

Or, dx  = $\frac{1}{6}$dy.

Or, I = $\mathop \smallint \nolimits \frac{{\frac{1}{6}.{\rm{dy}}}}{{\rm{y}}}$ = $\frac{1}{6}\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\rm{y}}}$ = $\frac{1}{6}$logy + c = $\frac{1}{6}$log(3x² – 4) + c.

 

                                                        

 

(iv) $\mathop \smallint \nolimits \frac{{{{\rm{x}}^{{\rm{n}} - 1}}}}{{\sqrt {{{\rm{a}}^{\rm{n}}} + {{\rm{x}}^{\rm{n}}}} }}$.dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{{{{\rm{x}}^{{\rm{n}} - 1}}}}{{\sqrt {{{\rm{a}}^{\rm{n}}} + {{\rm{x}}^{\rm{n}}}} }}$.dx

Put aⁿ + xⁿ  = y ⇒ n.x^{n – 1}.dx = dy ⇒ x^{n – 1}.dx = $\frac{{{\rm{dy}}}}{{\rm{n}}}$.

Then,                                                                                                                        

Or, I = $\frac{1}{{\rm{n}}}\mathop \smallint \nolimits {{\rm{y}}^{ - \frac{1}{2}}}$.dy = $\frac{1}{{\rm{n}}}\frac{{{{\rm{y}}^{ - \frac{1}{2} + 1}}}}{{\frac{1}{2}}}{\rm{\: \: }}$+ c

= $\frac{2}{{\rm{n}}}$y^1/2 + c

= $\frac{2}{{\rm{n}}}$$\sqrt {{{\rm{a}}^{\rm{n}}} + {{\rm{x}}^{\rm{n}}}} $ + c.

 

(v) $\mathop \smallint \nolimits \frac{{2{\rm{x}} + 3}}{{{{\left( {3{{\rm{x}}^2} + 9{\rm{x}} + 5} \right)}^3}}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{{2{\rm{x}} + 3}}{{{{\left( {3{{\rm{x}}^2} + 9{\rm{x}} + 5} \right)}^3}}}$.dx

Put 3x² + 9x + 5 = y

Or, (6x + 9).dx = dy

Or, 3(2x + 3).dx = dy

Or, (2x + 3).dx = $\frac{1}{3}$dy.

I = $\frac{1}{3}\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{{{\rm{y}}^3}}}$ = $\frac{1}{3}\mathop \smallint \nolimits {{\rm{y}}^{ - 3}}$.dy = $\frac{1}{3}\frac{{{{\rm{y}}^{ - 3 + 1}}}}{{ - 2}}$ + c.

= $ - \frac{1}{6}$.$\frac{1}{{{{\rm{y}}^2}}}$ + c

= $ - \frac{1}{{6{{\left( {3{{\rm{x}}^2} + 9{\rm{x}} + 5} \right)}^2}}}$ + c.

 

(vi) $\mathop \smallint \nolimits \frac{{\left( {3{\rm{x}} + 2} \right).{\rm{dx}}}}{{{{\left( {3{{\rm{x}}^2} + 4{\rm{x}} + 1} \right)}^3}}}$

Solution:

Let I = $\mathop \smallint \nolimits \frac{{\left( {3{\rm{x}} + 2} \right).{\rm{dx}}}}{{{{\left( {3{{\rm{x}}^2} + 4{\rm{x}} + 1} \right)}^3}}}$

Put 3x² + 4x + 1 = y

Or, (6x + 4).dx = dy

Or, 2(3x + 2).dx = dy

So, (3x + 2).dx = $\frac{1}{2}$.dy

I = $\mathop \smallint \nolimits \frac{{\left( {3{\rm{x}} + 2} \right).{\rm{dx}}}}{{{{\left( {3{{\rm{x}}^2} + 4{\rm{x}} + 1} \right)}^3}}}$ = $\frac{1}{2}\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{{{\rm{y}}^3}}}$.dy = $\frac{1}{2}\mathop \smallint \nolimits {{\rm{y}}^{ - 3}}$.dy

= $\frac{1}{2}.\frac{{{{\rm{y}}^{ - 3 + 1}}}}{{ - 2}}$ + c

= $ - \frac{1}{{4{{\rm{y}}^2}}}$ + c

= $ - \frac{1}{{4{{\left( {3{{\rm{x}}^2} + 4{\rm{x}} + 1} \right)}^2}}}$ + c.

 

(vii) $\mathop \smallint \nolimits 3{\rm{x}}\left( {{\rm{x}} + 2} \right){\left( {{{\rm{x}}^3} + 3{{\rm{x}}^2} + 1} \right)^2}$dx

Solution:

Let I = $\mathop \smallint \nolimits 3{\rm{x}}\left( {{\rm{x}} + 2} \right){\left( {{{\rm{x}}^3} + 3{{\rm{x}}^2} + 1} \right)^2}$dx

Put x³ + 3x² + 1 = y.

Or, (3x² + 6x).dx = dy

Or, 3x(x + 2)dx = dy.

I = $\mathop \smallint \nolimits {{\rm{y}}^2}.{\rm{dy}}$ = $\frac{{{{\rm{y}}^3}}}{3}$ + c = $\frac{1}{3}$(x³ + 3x² + 1)³ + c.

 

 

(viii) $\mathop \smallint \nolimits \frac{{{\rm{x}} + 2}}{{\sqrt {{{\rm{x}}^2} + 4{\rm{x}} + 3} }}$ dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{{{\rm{x}} + 2}}{{\sqrt {{{\rm{x}}^2} + 4{\rm{x}} + 3} }}$ dx

Put x³ + 4x + 3 = y.

Or, (2x + 4).dx = y

Or, 2(x + 2)dx = dy.

Or, (x + 2).dx = $\frac{1}{2}$dy

I = $\frac{1}{2}\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\sqrt {\rm{y}} }}$ = $\frac{1}{2}\mathop \smallint \nolimits {{\rm{y}}^{ - \frac{1}{2}}}.{\rm{dy}}$ = $\frac{1}{2}.\frac{{{{\rm{y}}^{ - \frac{1}{2} + 1}}}}{{\frac{1}{2}}}$ + c.

= ${{\rm{y}}^{\frac{1}{2}}}$ + c = $\sqrt {{{\rm{x}}^2} + 4{\rm{x}} + 3} $ + c.

 

 

(ix) $\mathop \smallint \nolimits \frac{{\left( {{{\rm{x}}^2} + 1} \right).{\rm{dx}}}}{{\sqrt {{{\rm{x}}^3} + 3{\rm{x}} + 3} }}$ dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{{\left( {{{\rm{x}}^2} + 1} \right).{\rm{dx}}}}{{\sqrt {{{\rm{x}}^3} + 3{\rm{x}} + 3} }}$ dx

= $\mathop \smallint \nolimits \frac{{\left( {{{\rm{x}}^2} + 1} \right).{\rm{dx}}}}{{{{\left( {{{\rm{x}}^3} + 3{\rm{x}} + 4} \right)}^{\frac{1}{2}}}.}}$dx

Put x³ + 3x + 4 = y.

Or, (3.x² + 3).dx = dy

Or, 3(x² + 1)dx = dy.

Or, (x² + 1).dx = $\frac{1}{3}$dy.

I = $\frac{1}{3}\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\sqrt {\rm{y}} }}$ = $\frac{1}{3}\mathop \smallint \nolimits {{\rm{y}}^{ - \frac{1}{2}}}.{\rm{dy}}$ = $\frac{1}{3}.\frac{{{{\rm{y}}^{ - \frac{1}{2} + 1}}}}{{\frac{1}{2}}}$ + c.

= $\frac{2}{3}$.y^1/2 + c = $\frac{2}{3}\sqrt {{{\rm{x}}^3} + 3{\rm{x}} + 4} $ + c.

 

2(i) $\mathop \smallint \nolimits {\rm{x}}.{\rm{sin}}\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{b}}} \right)$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\rm{x}}.{\rm{sin}}\left( {{\rm{a}}{{\rm{x}}^2} + {\rm{b}}} \right)$.dx

Put ax² + b = y

Or, 2ax.dx = y

So, x.dx = $\frac{1}{{2{\rm{a}}}}$dy

So, I = $\mathop \smallint \nolimits \frac{1}{{2{\rm{a}}}}.{\rm{siny}}.{\rm{dy}}$ = $\frac{1}{{2{\rm{a}}}}\mathop \smallint \nolimits {\rm{siny}}.$dy

= $\frac{1}{{2{\rm{a}}}}$(–cosy) + c = $ - \frac{1}{{2{\rm{a}}}}$cos(ax² + b) + c.

 

(ii) $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{cos}}\left( {{{\rm{x}}^3} + 1} \right)$.dx

Solution:

Let I = $\mathop \smallint \nolimits {{\rm{x}}^2}.{\rm{cos}}\left( {{{\rm{x}}^3} + 1} \right)$.dx

Put, x³ + 1 = y

Or, 3x².dx = dy

So, x³.dx = $\frac{1}{3}$dy

So, I = $\mathop \smallint \nolimits {\rm{cosy}}\frac{1}{3}$.dy  $\frac{1}{3}$siny + c = $\frac{1}{3}$sin(x³ + 1) + c.

 

(iii) $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}{\rm{sin}}\left( {{\rm{logx}}} \right)$.dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{1}{{\rm{x}}}{\rm{sin}}\left( {{\rm{logx}}} \right)$.dx

Put, logx = y

Or, $\frac{1}{{\rm{x}}}$.dx = dy

So, I = $\mathop \smallint \nolimits {\rm{siny}}$.dy

= – cosy + c = –cos(logx) + c.

 

(iv) $\mathop \smallint \nolimits {\sin ^3}{\rm{x}}.{\rm{cosx}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\sin ^3}{\rm{x}}.{\rm{cosx}}$.dx

Put, sinx = y

Or, cos xdx = dy.

So, I = $\mathop \smallint \nolimits {\rm{siny}}$.dy

= – cosy + c = –cos(logx) + c.

 

(v) $\mathop \smallint \nolimits {\sin ^2}{\rm{x}}.{\cos ^3}{\rm{x}}.$dx

Solution:

Let I = $\mathop \smallint \nolimits {\sin ^2}{\rm{x}}.{\cos ^3}{\rm{x}}.$dx

= $\mathop \smallint \nolimits {\sin ^2}{\rm{x}}.{\cos ^2}{\rm{x}}.{\rm{cosx}}.$dx = $\mathop \smallint \nolimits {\sin ^2}{\rm{x}}.\left( {1 - {{\sin }^2}{\rm{x}}} \right).{\rm{cosx}}$.dx

= $\mathop \smallint \nolimits ({\sin ^2}{\rm{x}} - {\sin ^4}{\rm{x}}).{\rm{cosx}}$.dx

Put, sinx = y

Or, cos xdx = dy.

So, I = $\mathop \smallint \nolimits \left( {{{\rm{y}}^2} - {{\rm{y}}^4}} \right)$dy = $\frac{{{{\rm{y}}^3}}}{3} - \frac{{{{\rm{y}}^5}}}{5}$ + c = $\frac{1}{3}$sin³x – $\frac{1}{5}$ sin⁵x + c

 

(vi) $\mathop \smallint \nolimits {\cos ^5}{\rm{x}}.{\sin ^3}{\rm{x}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\cos ^5}{\rm{x}}.{\sin ^3}{\rm{x}}$.dx

= $\mathop \smallint \nolimits {\cos ^5}{\rm{x}}.{\sin ^3}{\rm{x}}$dx = $\mathop \smallint \nolimits {\cos ^5}{\rm{x}}.{\sin ^2}{\rm{x}}.{\rm{sinx}}$.dx

= $\mathop \smallint \nolimits {\cos ^2}{\rm{x}}\left( {1 - {{\cos }^2}{\rm{x}}} \right).{\rm{sinx}}$.dx

= $\mathop \smallint \nolimits ({\cos ^2}{\rm{x}} - {\cos ^7}{\rm{x}}).{\rm{sinx}}$.dx

Put, cosx = y

Or, –sinx.dx = dy

So, sinx.dx = –dy

So, I = $\mathop \smallint \nolimits \left( {{{\rm{y}}^5} - {{\rm{y}}^2}} \right)( - $dy) = $ - \frac{{{{\rm{y}}^6}}}{6} + \frac{{{{\rm{y}}^8}}}{8}$ + c

= $\frac{1}{8}$y⁸ – $\frac{1}{6}$y⁶ + c = $\frac{1}{8}$cos⁸x – $\frac{1}{6}$cos⁶x + c.

 

(vii) $\mathop \smallint \nolimits {\left( {{\rm{asinx}} - {\rm{b}}} \right)^3}$.cosx.dx

Solution:

Let I = $\mathop \smallint \nolimits {\left( {{\rm{asinx}} - {\rm{b}}} \right)^3}$.cosx.dx

Put a sinx – b = y

Or, a.cosx.dx = dy

Or, cosx.dx = $\frac{1}{{\rm{a}}}$.dy

So, I = $\mathop \smallint \nolimits {{\rm{y}}^3}.\frac{1}{{\rm{a}}}$.dy = $\frac{1}{{\rm{a}}}\mathop \smallint \nolimits {{\rm{y}}^3}$.dy

= $\frac{1}{{\rm{a}}}$.$\frac{{{{\rm{y}}^4}}}{4}$ + c = $\frac{1}{{4{\rm{a}}}}$ (asinx – b)⁴ + c

 

(viii) $\mathop \smallint \nolimits \frac{{{\rm{sinx}}}}{{{{\left( {1 - {\rm{cosx}}} \right)}^{\rm{m}}}}}$.dx
Solution:

I = $\mathop \smallint \nolimits \frac{{{\rm{sinx}}}}{{{{\left( {1 - {\rm{cosx}}} \right)}^{\rm{m}}}}}$.dx

Put 1 – cosx = y

Or, – (–sinx).dx = dy

So, sinx.dx = dy.

So, I = $\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{{{\rm{y}}^{\rm{m}}}}}$ = $\mathop \smallint \nolimits {{\rm{y}}^{ - {\rm{m}}}}$.dy = $\frac{{{{\rm{y}}^{{\rm{m}} + 1}}}}{{ - {\rm{m}} + 1}}$ + c = $\frac{{{{\rm{y}}^{1 - {\rm{m}}}}}}{{1 - {\rm{m}}}}$ + c = $\frac{{{{\left( {1 - {\rm{cosx}}} \right)}^{1 - {\rm{m}}}}}}{{1 - {\rm{m}}}}$ + c.

 

(ix) $\mathop \smallint \nolimits {\rm{cotx}}{\left( {{\rm{logsinx}}} \right)^3}$.dx

Solution:

I = $\mathop \smallint \nolimits {\rm{cotx}}{\left( {{\rm{logsinx}}} \right)^3}$.dx

Put, log.sinx = y

Or, $\frac{1}{{{\rm{sinx}}}}$.cosx.dx = dy

So, cotx.dx = dy

So, I = $\mathop \smallint \nolimits {{\rm{y}}^3}$.dy = $\frac{{{{\rm{y}}^4}}}{4}$ + c = $\frac{1}{4}$(log.sinx)⁴ + c.

 

(x) $\mathop \smallint \nolimits {\tan ^2}\theta .{\sec ^4}\theta $.dθ = $\mathop \smallint \nolimits {\tan ^2}\theta .{\sec ^2}\theta .{\sec ^2}\theta $.dθ

Solution:

I = $\mathop \smallint \nolimits {\tan ^2}\theta .{\sec ^4}\theta $.dθ = $\mathop \smallint \nolimits {\tan ^2}\theta .{\sec ^2}\theta .{\sec ^2}\theta $.dθ

= $\mathop \smallint \nolimits {\tan ^2}\theta \left( {1 + {{\tan }^2}\theta } \right){\sec ^2}\theta $.dθ

Put, tanθ= y

Or, sec²θ.dθ = dy.

I = $\mathop \smallint \nolimits {{\rm{y}}^2}\left( {1 + {{\rm{y}}^2}} \right)$.dy = $\mathop \smallint \nolimits \left( {{{\rm{y}}^2} + {{\rm{y}}^4}} \right).{\rm{dy}}$

= $\frac{1}{3}$y³ + $\frac{1}{5}$y⁵ + c = $\frac{1}{3}$tan³θ + $\frac{1}{5}$tan⁵θ + c.

 

(xi) $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\sec ^4}{\rm{x}}$.dx = $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\sec ^2}{\rm{x}}.{\sec ^2}{\rm{x}}$.dx

Solution:

I = $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\sec ^4}{\rm{x}}$.dx = $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\sec ^2}{\rm{x}}.{\sec ^2}{\rm{x}}$.dx

= $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}\left( {1 + {{\tan }^2}{\rm{x}}} \right){\sec ^2}{\rm{x}}$.dx

Put, tanx= y

Or, sec²x.dx = dy.

I = $\mathop \smallint \nolimits {{\rm{y}}^3}\left( {1 + {{\rm{y}}^2}} \right)$.dy = $\mathop \smallint \nolimits \left( {{{\rm{y}}^3} + {{\rm{y}}^5}} \right).{\rm{dy}}$

= $\frac{{{{\rm{y}}^4}}}{4} + \frac{{{{\rm{y}}^6}}}{6} + {\rm{c}}$. = $\frac{1}{4}$tan⁴x + $\frac{1}{6}$tan⁶x + c.

 

(xii) $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\sec ^{\frac{3}{2}}}{\rm{x}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\sec ^{\frac{3}{2}}}{\rm{x}}$.dx

= $\mathop \smallint \nolimits {\tan ^2}{\rm{x}}.{\rm{tanx}}.{\sec ^{\frac{1}{2}}}{\rm{x}}.{\rm{secx}}$.dx

= $\mathop \smallint \nolimits ({\sec ^2}{\rm{x}} - 1).{\sec ^{\frac{1}{2}}}{\rm{x}}.{\rm{tanx}}.{\rm{secx}}$.dx

Put secx = y

Or, sec.x.tan.x.dx = dy

I = $\mathop \smallint \nolimits \left( {{{\rm{y}}^2} - 1} \right){{\rm{y}}^{\frac{1}{2}}}$.dy = $\mathop \smallint \nolimits \left( {{{\rm{y}}^{\frac{5}{2}}} - {{\rm{y}}^{\frac{1}{2}}}} \right)$.dy

= $\frac{{{{\rm{y}}^{\frac{7}{2}}}}}{{\frac{7}{2}}} - \frac{{{{\rm{y}}^{\frac{3}{2}}}}}{{\frac{3}{2}}}$ + c = $\frac{2}{7}$y^7/2 – $\frac{2}{3}$y^3/2 + c

= $\frac{2}{7}$sec^7/2 x – $\frac{2}{3}$sec^3/2x + c.

 

 

(xiii) $\mathop \smallint \nolimits {\cot ^{\frac{3}{2}}}{\rm{x}}.{\rm{cose}}{{\rm{c}}^4}{\rm{x}}.{\rm{dx}}$

Solution:

Let I = $\mathop \smallint \nolimits {\cot ^{\frac{3}{2}}}{\rm{x}}.{\rm{cose}}{{\rm{c}}^4}{\rm{x}}.{\rm{dx}}$

= $\mathop \smallint \nolimits {\cot ^{\frac{3}{2}}}{\rm{x}}.{\rm{cose}}{{\rm{c}}^2}{\rm{x}}.{\rm{cose}}{{\rm{c}}^2}{\rm{x}}$.dx

= $\mathop \smallint \nolimits {\cot ^{\frac{3}{2}}}{\rm{x}}\left( {1 + {{\cot }^2}{\rm{x}}} \right).{\rm{cose}}{{\rm{c}}^2}{\rm{x}}$.dx

Put cot x = y

Or, – cosec²xdx = dx

Or, cosec²x.dx = –dy

I = $ - \mathop \smallint \nolimits {{\rm{y}}^{\frac{3}{2}}}\left( {1 + {{\rm{y}}^2}} \right)$.dy = $ - \mathop \smallint \nolimits \left( {{{\rm{y}}^{\frac{3}{2}}} + {{\rm{y}}^{\frac{7}{2}}}} \right)$.dy

= $ - \frac{{{{\rm{y}}^{\frac{5}{2}}}}}{{\frac{5}{2}}} - \frac{{{{\rm{y}}^{\frac{9}{2}}}}}{{\frac{9}{2}}}$ + c

= $ - \frac{2}{5}$cot^5/2x – $\frac{2}{9}$.cot^9/2x + c.

 

 

(xiv) $\mathop \smallint \nolimits \frac{{{\rm{sinax}} + {\rm{cosax}}}}{{{\rm{sinax}} - {\rm{cosax}}}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{{{\rm{sinax}} + {\rm{cosax}}}}{{{\rm{sinax}} - {\rm{cosax}}}}$.dx

Put sinax – cosax = y

Or, {a.cosax – a.(–sinax)}dx = dy

Or, a(cosax + sinax)dx = dy

So, (cos ax + sin ax).dx = $\frac{1}{{\rm{a}}}$dy.

I = $\frac{1}{{\rm{a}}}\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\rm{y}}}$ = $\frac{1}{{\rm{a}}}$logy + c = $\frac{1}{{\rm{a}}}$log(sinax – cosax) + c.

 

 

(xv) $\mathop \smallint \nolimits {\rm{cotx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{cosx}}}}{{{\rm{sinx}}}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\rm{cotx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{cosx}}}}{{{\rm{sinx}}}}$.dx

Put sinx = y ⇒ cosx.dx = dy

I = $\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\rm{y}}}$ = log y + c = log(sinx) + c.

 

(xvi) $\mathop \smallint \nolimits {\rm{tanx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{secx}}.{\rm{tanx}}}}{{{\rm{secx}}}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\rm{tanx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{secx}}.{\rm{tanx}}}}{{{\rm{secx}}}}$.dx

Put sec.x = y ⇒ sec.x.tan.x.dx = dy

I = $\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\rm{y}}}$ = log y + c = log(secx) + c.

 

(xvii) $\mathop \smallint \nolimits {\rm{secx}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\rm{secx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{secx}}.\left( {{\rm{secx}} + {\rm{tanx}}} \right)}}{{{\rm{secx}} + {\rm{tanx}}}}$.dx

Put sec.x + tan.x = y

Or, (secx.tanx + sec²x).dx = dy

Or, secx(tanx + secx).dx = dy

I = $\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\rm{y}}}$ = log y + c = log(secx + tanx) + c.

 

(xviii) $\mathop \smallint \nolimits {\rm{cosecx}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\rm{cosecx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{cosecx}}.\left( {{\rm{cosecx}} - {\rm{cotx}}} \right)}}{{{\rm{cosecx}} - {\rm{cotx}}}}$.dx

Put, cosecx – cotx = y

Or, {–cosec x.cot.x – (–cosec²x)}.dx = dy

Or, cosecx(cosec.x – cot.x).dx = dy

I = $\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{\rm{y}}}$ = log y + c = log(cosecx– cotx) + c.

 

(xix) $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}$.dx
Solution:

Let I = $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}$.dx = $\mathop \smallint \nolimits {\rm{tanx}}.{\tan ^2}{\rm{x}}$.dx = $\mathop \smallint \nolimits {\rm{tanx}}({\sec ^2}{\rm{x}} - 1)$.dx

= $\mathop \smallint \nolimits ({\rm{tanx}}.{\sec ^2}{\rm{x}} - {\rm{tanx}})$.dx = $\mathop \smallint \nolimits {\rm{tanx}}.{\sec ^2}{\rm{dx}}$ – $\mathop \smallint \nolimits {\rm{tanx}}$.dx

= I₁ – I₂….(i)

I₁ = $\mathop \smallint \nolimits {\rm{tanx}}.{\sec ^2}{\rm{dx}}$

Put, tanx = y, sec²x.dx = dy.

Or, I₁ = $\mathop \smallint \nolimits {\rm{y}}$.dy = $\frac{1}{2}$y² + c₁ = $\frac{1}{2}$tan²x + c₁

Or, I₂ = $\mathop \smallint \nolimits {\rm{tanx}}.$dx = $\mathop \smallint \nolimits \frac{{{\rm{sinx}}}}{{{\rm{cosx}}}}$.dx

Put cosx = z

Or, –sinx.dx = dz

Or, sinx.dx = –dz

I₂ = $\mathop \smallint \nolimits  - \frac{{{\rm{dz}}}}{{\rm{z}}}$ = – logz + c₂ = – log(cosx) + c₂

From (i), I = I₁ – I₂.

= $\frac{1}{2}$tan²x + c₁ – {–log(cosx) + c₂}

= $\frac{1}{2}$tan²x + log.cosx + c₁ + c₂.

= $\frac{1}{2}$tan²x + log.cosx + c, where c = c₁ – c₂.

 

(xx) $\mathop \smallint \nolimits {\cot ^3}{\rm{x}}$.dx

Solution:

Let I = $\mathop \smallint \nolimits {\cot ^3}{\rm{x}}$.dx = $\mathop \smallint \nolimits {\rm{cotx}}.{\cot ^2}{\rm{x}}$.dx = $\mathop \smallint \nolimits {\rm{cotx}}({\rm{cose}}{{\rm{c}}^2}{\rm{x}} - 1)$.dx

= $\mathop \smallint \nolimits ({\rm{cotx}}.{\rm{cose}}{{\rm{c}}^2}{\rm{x}} - {\rm{cotx}})$.dx = $\mathop \smallint \nolimits {\rm{cotx}}.{\rm{cose}}{{\rm{c}}^2}{\rm{dx}}$ – $\mathop \smallint \nolimits {\rm{cotx}}$.dx

I₁ = $\mathop \smallint \nolimits {\rm{cotx}}.{\rm{cose}}{{\rm{c}}^2}{\rm{dx\: }}$

= I₁ – I₂….(i)

I₁ = $ - \mathop \smallint \nolimits {\rm{y}}$.dy = $ - \frac{1}{2}$y² + c₁ = $ - \frac{1}{2}$cot²x + c₁.

I₂ = $\mathop \smallint \nolimits {\rm{cotx}}.$dx = $\mathop \smallint \nolimits \frac{{{\rm{cosx}}}}{{{\rm{sinx}}}}$.dx

Put, sinx = z

Or, cosx.dx = dz

Or, I₂ = $\mathop \smallint \nolimits \frac{{{\rm{dz}}}}{{\rm{z}}}$ = logz + c₂ = log(sin x) + c₂.

From (i), I = I₁– I₂

= $ - \frac{1}{2}$cot²x + c₁ – {log(sinx) + c₂}

= $ - \frac{1}{2}$cot²x – log(sinx) + c₁ + c₂.

= $ - \frac{1}{2}$ cot²x – log(sinx) + c, c = c₁ – c₂.

 

(xxi) $\mathop \smallint \nolimits {\sec ^4}{\rm{x}}.{\rm{dx}}$

Solution:

Let I = $\mathop \smallint \nolimits {\sec ^4}{\rm{x}}.{\rm{dx}}$ = $\mathop \smallint \nolimits {\sec ^2}{\rm{x}}.{\sec ^2}{\rm{x}}$.dx

= $\mathop \smallint \nolimits (1 + {\tan ^2}{\rm{x}}){\sec ^2}{\rm{x}}$.dx

Put tanx = y

Or, sec²x.dx = dy

I = $\mathop \smallint \nolimits \left( {1 + {{\rm{y}}^2}} \right)$.dy = y + $\frac{1}{3}$y³ + c = tanx + $\frac{1}{3}$tan³x + c.

 

(xxii) $\mathop \smallint \nolimits {\tan ^4}{\rm{x}}.{\rm{dx}}$

Solution:

Let I = $\mathop \smallint \nolimits {\tan ^4}{\rm{x}}.{\rm{dx}}$ = $\mathop \smallint \nolimits {\tan ^2}{\rm{x}}.{\tan ^2}{\rm{x}}$.dx

= $\mathop \smallint \nolimits ({\sec ^2}{\rm{x}} - 1){\tan ^2}{\rm{x}}$.dx = $\mathop \smallint \nolimits ({\sec ^2}{\rm{x}}.{\tan ^2}{\rm{x}} - {\tan ^2}{\rm{x}})$.dx

= $\mathop \smallint \nolimits \{ {\sec ^2}{\rm{x}}.{\tan ^2}{\rm{x}} - ({\sec ^2}{\rm{x}} - 1)\} $.dx

= $\mathop \smallint \nolimits \{ {\sec ^2}{\rm{x}}.{\tan ^2}{\rm{x}} - {\sec ^2}{\rm{x}} + 1\} $.dx

= $\mathop \smallint \nolimits ({\tan ^2}{\rm{x}} - 1){\sec ^2}{\rm{x}}$.dx + $\mathop \smallint \nolimits 1$.dx

Put tanx = y

Or, sec²x.dx = dy

I = $\mathop \smallint \nolimits \left( {{{\rm{y}}^2} + 1} \right)$.dy + $\mathop \smallint \nolimits 1.{\rm{dx}}$

= $\frac{1}{3}$y³ – y + x + c.

= $\frac{1}{3}$tan³x – tanx + x + c.

 

(xxiii) $\mathop \smallint \nolimits {\tan ^5}{\rm{x}}.{\rm{dx}}$

Solution:

Let I = $\mathop \smallint \nolimits {\tan ^5}{\rm{x}}.{\rm{dx}}$ = $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}.{\tan ^2}{\rm{x}}$.dx

= $\mathop \smallint \nolimits {\tan ^3}{\rm{x}}({\sec ^2}{\rm{x}} - 1)$.dx = $\mathop \smallint \nolimits ({\tan ^3}{\rm{x}}.{\sec ^2}{\rm{x}} - {\tan ^3}{\rm{x}})$.dx

= $\mathop \smallint \nolimits ({\tan ^3}{\rm{x}}.{\sec ^2}{\rm{x}} - {\rm{tanx}}.{\tan ^2}{\rm{x}})$.dx

= $\mathop \smallint \nolimits \{ {\tan ^3}{\rm{x}}.{\sec ^2}{\rm{x}} - {\rm{tanx}}\left( {{{\sec }^2}{\rm{x}} - 1} \right)\} $.dx

= $\mathop \smallint \nolimits \{ ({\tan ^3}{\rm{x}}.{\sec ^2}{\rm{x}} - {\rm{tanx}}.{\sec ^2}{\rm{x}} + {\rm{tanx}})$.dx

= $\mathop \smallint \nolimits \{ ({\tan ^3}{\rm{x}} - {\rm{tanx}}).{\sec ^2}{\rm{x}} + {\rm{tanx}}\} .$dx

= $\mathop \smallint \nolimits ({\tan ^3}{\rm{x}} - {\rm{tanx}}){\sec ^2}{\rm{x}}$.dx + $\mathop \smallint \nolimits {\rm{tanx}}$.dx

= I₁ – I₂ ….(i)

I₁ = $\mathop \smallint \nolimits \left( {{{\rm{y}}^3} - {\rm{y}}} \right)$.dy = $\frac{1}{4}$.y⁴ – $\frac{1}{2}$y² + c₁ = $\frac{1}{4}$tan⁴x – $\frac{1}{2}$tan²x + c₁

I₂ = $\mathop \smallint \nolimits {\rm{tanx}}$.dx = $\mathop \smallint \nolimits \frac{{{\rm{tanx}}.{\rm{secx}}}}{{{\rm{secx}}}}$.dx

Put sec x = z, secx.tanx.dx = dz

I₂ = $\mathop \smallint \nolimits \frac{{{\rm{dz}}}}{{\rm{z}}}$ = logz + c₂ = log(secx) + c₂

From(i), I = I₁ + I₂

= $\frac{1}{4}$tan⁴x – $\frac{1}{2}$tan²x + c₁ + {log(secx) + c₂}

= $\frac{1}{4}$tan⁴x – $\frac{1}{2}$tan²x + log(secx) + c₁ + c₂

= $\frac{1}{4}$tan⁴x – $\frac{1}{2}$ tan²x + log(secx) + c, c₁ + c₂.

 

(xxiv) $\mathop \smallint \nolimits ({\tan ^2}{\rm{x}} + {\tan ^4}{\rm{x}})$.dx

Solution:

Let I = $\mathop \smallint \nolimits ({\tan ^2}{\rm{x}} + {\tan ^4}{\rm{x}})$.dx = $\mathop \smallint \nolimits {\tan ^2}{\rm{x}}$(1 + tan²x).dx

= $\mathop \smallint \nolimits ({\tan ^2}{\rm{x}}.{\sec ^2}{\rm{x}}.{\rm{dx}}$

Put tanx = y

Or, sec²x.dx = dy

Put tanx = y

Or, I = $\mathop \smallint \nolimits {{\rm{y}}^2}$.dy = $\frac{1}{3}$y³ + c = $\frac{1}{3}$ tan³x + c.

 

(xxv) $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{sinx}}.{\rm{cosx}}}}$.cos2x.dx

Solution:

Let I = $\mathop \smallint \nolimits {{\rm{e}}^{{\rm{sinx}}.{\rm{cosx}}}}$.cos2x.dx

Put sinx.cosx = y.

Or, $\frac{1}{2}$.2sinx.cosx = y

Or, $\frac{1}{2}$sin2x = y

Or, $\frac{1}{2}$cos2x.2dx = dy

Or, cos2x.dx = dy

I = $\mathop \smallint \nolimits {{\rm{e}}^{\rm{y}}}$.dy = e^y + c = e^sinx.cosx + c.

 

(xxvi) $\mathop \smallint \nolimits {{\rm{e}}^{{{\sin }^2}{\rm{x}}}}$.sin2x.dx

Solution:

Let I = $\mathop \smallint \nolimits {{\rm{e}}^{{{\sin }^2}{\rm{x}}}}$.sin2x.dx

Put sin²x = y or, 2sinx.cosx.dy = dy

Or, sin2x.dx = dy

I = $\mathop \smallint \nolimits {{\rm{e}}^{\rm{y}}}$.dy = e^y + c = e^sin2x + c.

 

(xxvii) $\mathop \smallint \nolimits \left( {1 - \frac{1}{{{{\rm{x}}^2}}}} \right)$.e^{x + 1/x}.dx

Solution:

Let I = $\mathop \smallint \nolimits \left( {1 - \frac{1}{{{{\rm{x}}^2}}}} \right)$.e^{x + 1/x}.dx

Put x + $\frac{1}{{\rm{x}}}$ = y.

Or, $\left( {1 - \frac{1}{{{{\rm{x}}^2}}}} \right)$.dx = dy

I = $\mathop \smallint \nolimits {{\rm{e}}^{\rm{y}}}$.dy = e^y + c = e^{x + 1/x} + c.

 

(xxviii) $\mathop \smallint \nolimits \frac{{{\rm{sin}}\sqrt {\rm{x}} }}{{\sqrt {\rm{x}} }}$.dx

Solution:

Let I = $\mathop \smallint \nolimits \frac{{{\rm{sin}}\sqrt {\rm{x}} }}{{\sqrt {\rm{x}} }}$.dx

Put, $\sqrt {\rm{x}} $ = y

Or, $\frac{1}{{2\sqrt {\rm{x}} }}$.dx = dy

Or, $\frac{1}{{\sqrt {\rm{x}} }}$.dx = 2dy.

I = $\mathop \smallint \nolimits {\rm{siny}}.2{\rm{dy}}$ = 2$\mathop \smallint \nolimits {\rm{siny}}$.dy

= 2(–cosy) + c = – 2cos$\sqrt {\rm{x}} $ + c.

 

(xxix) $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{2{\rm{x}}}}}}{{1 + {{\rm{e}}^{\rm{x}}}}}$.dx

Solution:

Or, Let I = $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{2{\rm{x}}}}}}{{1 + {{\rm{e}}^{\rm{x}}}}}$.dx = $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{\rm{x}}}.{{\rm{e}}^{\rm{x}}}}}{{1 + {{\rm{e}}^{\rm{x}}}}}$.dx

Put e^x = y

Or, e^x.dx = dy

I = $\mathop \smallint \nolimits \frac{{{\rm{ydy}}}}{{1 + {\rm{y}}}}$ = $\mathop \smallint \nolimits \frac{{\left\{ {\left( {1 + {\rm{y}}} \right) - 1} \right\}}}{{1 + {\rm{y}}}}$.dy

= $\mathop \smallint \nolimits \left( {1 - \frac{1}{{1 + {\rm{y}}}}} \right)$.dy = y – log (1 + y) + c = e^x – log(1 + e^x) + c.

 

(xxx) $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{\rm{x}}} - 1}}{{{{\rm{e}}^{\rm{x}}} + 1}}$.dx

Solution:

Or, Let I = $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{\rm{x}}} - 1}}{{{{\rm{e}}^{\rm{x}}} + 1}}$.dx = $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{ - \frac{{\rm{x}}}{2}}}\left( {{{\rm{e}}^{\rm{x}}} - 1} \right)}}{{{{\rm{e}}^{ - \frac{{\rm{x}}}{2}}}\left( {{{\rm{e}}^{\rm{x}}} + 1} \right)}}$.dx

= $\mathop \smallint \nolimits \frac{{{{\rm{e}}^{\frac{{\rm{x}}}{2}}} - {{\rm{e}}^{ - \frac{{\rm{x}}}{2}}}}}{{{{\rm{e}}^{\frac{{\rm{x}}}{2}}} + {{\rm{e}}^{ - \frac{{\rm{x}}}{2}}}}}$

Put e^x/2 + e^–x/2 = y.

Or, $\left\{ {{{\rm{e}}^{\frac{{\rm{x}}}{2}}}.\frac{1}{2} + {{\rm{e}}^{ - \frac{{\rm{x}}}{2}}}\left( { - \frac{1}{2}} \right)} \right\}$.dx = dy

Or, $\frac{1}{2}$ (e^x/2 – e^–x/2).dx = dy

Or, (e^x/2 – e^–x/2).dx = 2.dy

I = $\mathop \smallint \nolimits \frac{{2.{\rm{dy}}}}{{\rm{y}}}$ = 2.logy + c = 2log(e^x/2 + e^–x/2) + c.

 

(xxxi) $\mathop \smallint \nolimits \frac{{{\rm{x}}{{\rm{e}}^{\rm{x}}}.{\rm{dx}}}}{{{{\cos }^2}({\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}}}$

Solution:

Let I = $\mathop \smallint \nolimits \frac{{{\rm{x}}{{\rm{e}}^{\rm{x}}}.{\rm{dx}}}}{{{{\cos }^2}({\rm{x}}{{\rm{e}}^{\rm{x}}} - {{\rm{e}}^{\rm{x}}}}}$

Put x.e^x – e^x = y.

Or, {x.e^x + e^x.1 – e^x}dx = dy

Or, x.e^x.dx = dy.

I = $\mathop \smallint \nolimits \frac{{{\rm{dy}}}}{{{{\cos }^2}{\rm{y}}}}$ = $\mathop \smallint \nolimits {\sec ^2}{\rm{y}}.{\rm{dy}}$ = tany + c = tan(xe^x – e^x) + c