Consider the integrals A=∫_0^π ( x d x)/( x+ x) and B=∫_0^π ( x d x)/( x- x) What is the value of B ?

Correct answer: B. (π)/(2) aligned & B=∫_0^π ( x)/( x- x) × (( x+ x))/(( x+ x)) d x & =-∫_0^π ( ^2 x+ x x)/( 2 x) d x & =(-1)/(2) ∫_0^π (2 ^2 x)/( 2 x) d x-(1)/(2) ∫_0^π (2 x x)/( 2 x) d x & =(-1)/(2) ∫_0^π (1- 2 x)/( 2 x) d x-(1)/(2) ∫_0^π 2 x d x & =(-1)/(2) ∫_0^π 2 x d x+(1)/(2) ∫_0^π d x-(1)/(2) ∫_0^π 2 x d x & =(-1)/(2)[( | 2 x+ 2 x|)/(2)]_0^π+(1)/(2)[x]_0^π-(1)/(2)[( | 2 x|)/(2)]_0^π & = & (-1)/(4)[ (1+0)- (1+0)]+(1)/(2)[π+0]-(1)/(4)[ (1)- (1)] aligned aligned & = & (-1)/(4)[ (1+0)- (1+0)]+(1)/(2)[π+0]-(1)/(4)[ (1)- (1)] & =0+π / 2-0=π / 2 aligned

NDA 2015

Exam: NDA
Year: 2015
Subject: Mathematics
Chapter: Definite Integration

NDA 2015 · Mathematics · Definite Integration

Consider the integrals A=∫_0^π ( x d x)/( x+ x) and B=∫_0^π ( x d...

Consider the integrals A=\int_0^\pi

\frac{\sin x

d x}{

\sin x+\cos

\text{

and } B=\int_0^\pi

\frac{\sin x

d x}{

\sin x-\cos

x} What is the value of

B

Options

A $\frac{\pi}{4}$
B $\frac{\pi}{2}$
C $\frac{3 \pi}{4}$
D $\pi$

Correct answer

B. $\frac{\pi}{2}$

Step-by-step solution

\begin{aligned} & B=\int_0^\pi \frac{\sin x}{\sin x-\cos x} \times \frac{(\sin x+\cos x)}{(\sin x+\cos x)} d x \\ & =-\int_0^\pi \frac{\sin ^2 x+\sin x \cos x}{\cos 2 x} d x \\ & =\frac{-1}{2} \int_0^\pi \frac{2 \sin ^2 x}{\cos 2 x} d x-\frac{1}{2} \int_0^\pi \frac{2 \sin x \cos x}{\cos 2 x} d x \\ & =\frac{-1}{2} \int_0^\pi \frac{1-\cos 2 x}{\cos 2 x} d x-\frac{1}{2} \int_0^\pi \tan 2 x d x \\ & =\frac{-1}{2} \int_0^\pi \sec 2 x d x+\frac{1}{2} \int_0^\pi d x-\frac{1}{2} \int_0^\pi \tan 2 x d x \\ & =\frac{-1}{2}\left[\frac{\log |\sec 2 x+\tan 2 x|}{2}\right]_0^\pi+\frac{1}{2}[x]_0^\pi-\frac{1}{2}\left[\frac{\log |\sec 2 x|}{2}\right]_0^\pi \\ & = \\ & \frac{-1}{4}[\log (1+0)-\log (1+0)]+\frac{1}{2}[\pi+0]-\frac{1}{4}[\log (1)-\log (1)]\end{aligned}

\begin{aligned} & = \\ & \frac{-1}{4}[\log (1+0)-\log (1+0)]+\frac{1}{2}[\pi+0]-\frac{1}{4}[\log (1)-\log (1)] \\ & =0+\pi / 2-0=\pi / 2\end{aligned}

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