Integration using Trigonometric Identities MCQ Quiz - Objective Question with Answer for Integration using Trigonometric Identities - Download Free PDF

Calculation

∵ \(\frac{d}{d x}\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}\right)=\frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{3 / 2}}+\frac{x}{1-x^{2}}\)

⇒ \(\int e^{x}\left(\frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}+\frac{\sin ^{-1} x}{\left(1-x^{2}\right)^{3 / 2}}+\frac{x}{1-x^{2}}\right) d x\)

= \(e^{x} \cdot \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}}+c=g(x)+C\)

Note : assuming g(x) = \(\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}\)

\(g(1 / 2)=\frac{\mathrm{e}^{1 / 2}}{2} \cdot \frac{\frac{\pi}{6} \times 2}{\sqrt{3}}=\frac{\pi}{6} \sqrt{\frac{\mathrm{e}}{3}}\)

Comment : In this question we will not get a unique function g(x), but in order to match the answer we will have to assume g(x) = \(\frac{\mathrm{xe}^{\mathrm{x}} \sin ^{-1} \mathrm{x}}{\sqrt{1-\mathrm{x}^{2}}}\).

Hence option 3 is correct

Calculation

Given integral: ∫ \(\frac{2\cos 2x}{(1+\sin 2x)(1+\cos 2x)}\) dx

⇒ ∫ \(\frac{2(1 - 2\sin^2 x)}{(1+2\sin x \cos x)(1+2\cos^2 x - 1)}\) dx

⇒ ∫ \(\frac{2(1 - 2\sin^2 x)}{(1+2\sin x \cos x)(2\cos^2 x)}\) dx

⇒ ∫ \(\frac{1 - 2\sin^2 x}{\cos^2 x (1+2\sin x \cos x)}\) dx

⇒ ∫ \(\frac{\cos^2 x - \sin^2 x}{\cos^2 x (1+2\sin x \cos x)}\) dx

⇒ ∫ \(\frac{\cos^2 x - \sin^2 x}{\cos^2 x ((\sin x + \cos x)^2)}\) dx

⇒ ∫ \(\frac{\cos^2 x - \sin^2 x}{\cos^2 x (\cos^2 x (1+\tan x)^2)}\) dx

⇒ ∫ \(\frac{\cos^2 x - \sin^2 x}{\cos^4 x (1+\tan x)^2}\) dx

⇒ ∫ \(\frac{1 - \tan^2 x}{\cos^2 x (1+\tan x)^2}\) dx

⇒ ∫ \(\frac{(1 - \tan x)(1 + \tan x)}{\cos^2 x (1+\tan x)^2}\) dx

⇒ ∫ \(\frac{1 - \tan x}{\cos^2 x (1+\tan x)}\) dx

⇒ ∫ \(\frac{\sec^2 x (1 - \tan x)}{(1+\tan x)}\) dx

Let t = 1 + tan x, then dt = sec² x dx

⇒ ∫ \(\frac{1 - (t - 1)}{t}\) dt

⇒ ∫ \(\frac{2 - t}{t}\) dt

⇒ ∫ (\(\frac{2}{t} - 1\)) dt

⇒ 2 ln |t| - t + c

⇒ 2 ln |1 + tan x| - (1 + tan x) + c

⇒ 2 ln |1 + tan x| - tan x - 1 + c

⇒ 2 ln |1 + tan x| - tan x + C (where C = c - 1)

∴ The integral is 2 ln |1 + tan x| - tan x + C

Hence option 4 is correct

Calculation

Let \(I = \int \frac{2x^2 \cos(x^2) - \sin(x^2)}{x^2} dx\)

\(I = \int \left( \frac{2x^2 \cos(x^2)}{x^2} - \frac{\sin(x^2)}{x^2} \right) dx\)

\(I = \int 2 \cos(x^2) dx - \int \frac{\sin(x^2)}{x^2} dx\)

Let's consider the derivative of \(\frac{\sin(x^2)}{x}\):

\(\frac{d}{dx} \left( \frac{\sin(x^2)}{x} \right) = \frac{2x^2 \cos(x^2) - \sin(x^2)}{x^2}\)

Therefore, \(I = \int \frac{d}{dx} \left( \frac{\sin(x^2)}{x} \right) dx\)

⇒ \(I = \frac{\sin(x^2)}{x} + C\)

∴ The integral is \(\frac{\sin(x^2)}{x} + C\)

Hence option 4 is correct

Calculation:

\(\int e^{x} \frac{(1+\sin x)}{(1+\cos x)} \mathrm{dx}\)

= \(\int e^{x}\left[\frac{1}{2} \sec ^{2} \frac{x}{2}+\tan \frac{x}{2}\right] \mathrm{dx}\)

= \(\frac{1}{2} \int e^{x} \sec ^{2} \frac{x}{2} \mathrm{dx}+\int e^{x} \tan \frac{x}{2} \mathrm{dx}\)

= \(e^{x} \tan \frac{x}{2}+C\)

But I = e^xf(x) + C (given)

∴ \(f(x)=\tan \frac{x}{2}\)

Hence option 3 is correct

Calculation 

\(\int \frac{sin^2 x - cos^2 x}{sin^2 x cos^2 x} dx = \int (\frac{sin^2 x}{sin^2 x cos^2 x} - \frac{cos^2 x}{sin^2 x cos^2 x}) dx\)

⇒ \(\int (sec^2x - cosec^2x) dx\)

⇒ tan x + cot x + c

Hence option 1 is correct

Concept:

1 + cos 2x = 2cos² x

1 - cos 2x = 2sin² x

\(\rm \int \cos x\;dx = \sin x + c\)

Calculation:

I = \(\rm \int cos^2 x\;dx\)

= \(\rm \int \frac{1+\cos 2x}{2}\;dx\)

= \(\rm \frac{1}{2}\int (1+\cos 2x)\;dx\)

= \(\rm \frac{1}{2} \left[x+\frac{\sin 2x}{2} \right ] + c\)

= \(\rm \frac{x}{2}+\frac{\sin 2x}{4} + c\)

Concept:

1 - cos 2x = 2 sin² x

1 – sin² x = cos² x

\(\smallint {\sec ^2}{\rm{xdx}} = \tan {\rm{x}} + {\rm{c}}\)

Calculation:

Let I = \(\smallint \frac{{1 - \cos 2{\rm{x}}}}{{1 - {{\sin }^2}{\rm{x}}}}{\rm{\;dx}}\)

\( = \smallint \frac{{2{{\sin }^2}{\rm{x}}}}{{{{\cos }^2}{\rm{x}}}}{\rm{\;dx}}\)

\( = 2\smallint {\tan ^2}{\rm{xdx}}\)

\( = 2\smallint \left( {{{\sec }^2}{\rm{x}} - 1} \right){\rm{dx}}\)

= 2 [tan x – x] + c

= 2 tan x – 2x + c

Concept:

• \(\int x^n dx = \frac{x^{n+1}}{n+1} + C, n \neq -1\)

• \(\frac{d}{dx}\sec x=\sec x\, \tan x\)

Calculation:

We have,

∫ secn x tan x dx

⇒ ∫ sec ^{n - 1} x (sec x tan x) dx      ----(1)

Let sec x = t

⇒ sec x tan x dx = dt

On substituting these values in equation (1), we get

∫ t ^{n - 1} dt 

\(\Rightarrow \frac{t^n}{n}+C\)

\(\Rightarrow \frac{1}{n}\sec ^n x+C\)

Hence, ∫ secn x tan x dx = \(\frac{1}{n}\sec ^n x+C\)

Concept:

sin²x + cos²x = 1

\(\rm \int sin x dx = -cosx \)

\(\rm \int cos x dx = sin x \)

Calculation:

We have \(\rm \int \sqrt{1 - sin 2x} \space dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{sin^{2}x + cos^{2}x - 2sinx cosx} )dx\) = A sinx + B cosx + C

⇒ \(\rm \int (\sqrt{(sinx - cosx)^{2}})dx\) = A sinx + B cosx + C

⇒ \(\rm \int (|(sinx - cosx)|)dx\) = A sinx + B cosx + C     ----(i)

If 0 < x < \(\frac{\pi}{4}\), then sinx < cosx

⇒ |sinx - cosx| = -sinx + cosx    -----(ii)

Now from (i) and (ii), we get 

⇒ \(\rm \int (-\space sinx + cosx) \space dx\) = A sinx + B cosx + C

⇒ cosx + sinx + C = A sinx + B cosx + C

On comparing A = 1, B = 1 and C = 0

Hence, A + B - 2 = 0 is correct.

Concept:

\( \rm \int x^n dx = \frac{x^{n+1}}{n+1}+c\)

Calculation:

To Find: Integral of sec² x with respect to sec x

\(\rm \int \sec^2 x\; d(\sec x)\)

Replace sec x = t, we get

\(= \rm \int t^2 dt\\= \frac{t^3}{3}+c\\=\frac{\sec^3 x}{3}+c\)

Concept:

180° = π radian

\(\smallint \sin {\rm{xdx}} = {\rm{\;}} - \cos {\rm{x}} + {\rm{c}}\)

Calculation:

As we know that, 180° = π radian

∴ 1° = \(\frac{{\rm{\pi }}}{{180}}\) radian

So, x° = \(\frac{{\rm{\pi x }}}{{180}}\) radian

Let I = \(\smallint \sin {\rm{x}}^\circ {\rm{dx}}\)

\( = \smallint \sin \frac{{{\rm{\pi x}}}}{{180}}{\rm{dx}}\)

Let \(\frac{{\rm{\pi x }}}{{180}}\)= t

⇒ dx = \(\frac {180}{\pi}\) dt

\({\rm{I}} = \frac{{180}}{{\rm{\pi }}}\smallint \sin {\rm{tdt}} = {\rm{\;}} - \frac{{180\cos {\rm{t}}}}{{\rm{\pi }}} + {\rm{c}} = {\rm{\;}} - \frac{{180\cos {\rm{x}}^\circ }}{{\rm{\pi }}} + {\rm{c}}\)

Concept:

\(\rm \int \sec^{2}xdx = \tan x + c \\ \rm \int cosec^{2}xdx = -\cot x+c\)

Calculation:

Let I = \(\rm \int{{\left( \frac {1}{\cos^2 x} - \frac {1}{\sin^2 x} \right)}} dx\)

\(= \rm \int \sec^{2}xdx - \rm \int cosec^{2}xdx\)

= tan x - (-cot x) + c

\(= \rm \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x}+c \\ =\frac{\sin^2 x+ \cos^2 x}{\sin x \cos x}+c \\ = \frac{1}{\sin x \cos x}+c\)

\(= \rm \frac{2}{2\sin x \cos x} +c\\= \frac{2}{\sin2x}+c\)            (∵ 2 sin x cos x = sin 2x)

= 2 cosec 2x + c

Formula:

\(\int {e}^x [f(x) + f'(x)] dx = e^x f(x) + C\)

Calculation:

Let \(I = \int e^x . \frac{1 + sinx}{1 + cos x} dx\)

⇒ \(I = \int e^x . \frac{1 + 2sin\frac{x}{2}cos \frac{x}{2}}{ 2cos^2 \frac{x}{2}} dx\)

⇒ \(I = \int e^x . [\frac{1}{2 cos^2\frac{x}{2}} + \frac{2 sin\frac{x}{2} cos\frac{x}{2}}{2 cos^2\frac{x}{2}}] dx\)

⇒ \(I = \int e^x . [\frac{1}{2} sec^2 \frac{x}{2} + tan \frac{x}{2}] dx\)

The above integrand is of the form \(\int {e}^x [f(x) + f'(x)] dx\)

\(f(x) = tan \frac{x}{2}\)

 \(f'(x) = \frac{1}{2} sec^2 \frac{x}{2}\)

 \(∴\ I = e^x f(x) + c\)

\(I = e^x tan \frac{x}{2} + c\)

Concept:

1 + cos 2x = 2 cos² x

1 – cos² x = sin² x

\(\smallint {\rm{cose}}{{\rm{c}}^2}{\rm{xdx}} = - \cot {\rm{x}} + {\rm{c}}\)

Calculation:

Let I = \(\smallint \frac{{1 + \cos 2{\rm{x}}}}{{1 - {{\cos }^2}{\rm{x}}}}{\rm{\;dx}}\)    ---- (∵ 1 + cos 2x = 2 cos2 x and 1 – cos2 x = sin2 x)

\(= \smallint \frac{{2{{\cos }^2}{\rm{x}}}}{{{{\sin }^2}{\rm{x}}}}{\rm{\;dx}}\)

\(= 2\smallint {\cot ^2}{\rm{xdx}}\)

\( = 2\smallint \left( {{\rm{cose}}{{\rm{c}}^2}{\rm{x}} - 1} \right){\rm{dx}}\)

= 2 [-cot x – x] + c

= -2 cot x – 2x + c

Concept:

Some useful formulas are:

∫cosx dx = sinx + c

∫ secx dx = ln(sec x + tan x) + c

cos2θ = 2cos²θ - 1

\(\rm 1\over cos θ \)= secθ 

Calculation:

Given integration is, \(\rm ∫{ {cos2x}\over {cosx} }dx\)

= \(\rm ∫{ {2cos^2x-1}\over {cosx} }dx\)

= \(\rm ∫({ {{2cos^2x}\over {cosx}}-{{1}\over {cosx} }})dx\)

= \(\rm ∫({ {{2cosx}}-{secx} })dx\)

= 2sinx - ln(sec x + tan x) + C, C = constant of integration