Paragraph: Let I=∫_0^ (π)/(2) (f(x))/(g(x)) d x , where f(x)= x and g(x)= x+ x+1 . Question: What is ∫_0^ (π)/(2) (d x)/(g(x)) equal to?

Correct answer: C. 2 Given that g(x)= x+ x+1 aligned & ∫_0^ (π)/(2) (d x)/(g(x))=∫_0^ (π)/(2) (1)/( x+ x+1) d x & =∫_0^ (π)/(2) (1)/((2 (π)/(2))/(1+ ^2 (π)/(2))+(1- ^2 (π)/(2))/(1+ ^2 (π)/(2))+1) d x aligned aligned & =∫_0^ (π)/(2) (1+ ^2 (x)/(2))/(2 (x)/(2)+1- ^2 (x)/(2)+1+ ^2 (x)/(2)) d x & =∫_0^ (π)/(2) ( ^2 (x)/(2))/(2(1+ (x)/(2))) d x & let 1+ (x)/(2)=t (1)/(2) ^2 (x)/(2) d x=d t aligned when x=π / 2 then t=2 , and when x=0 then t=1=∫_1^2 1 t dt=[ t]_1^2= 2- 1= 2

NDA 2024

Exam: NDA
Year: 2024
Subject: Mathematics
Chapter: Definite Integration

NDA 2024 · Mathematics · Definite Integration

Paragraph: Let I=∫_0^ (π)/(2) (f(x))/(g(x)) d x , where f(x)= x and...

Paragraph: Let

\mathrm{I}=\int_0^{\frac{\pi}{2}} \frac{f(x)}{g(x)} d x

, where

f(x)=\sin x

and

g(x)=\sin x+\cos x+1

.
Question: What is

\int_0^{\frac{\pi}{2}} \frac{d x}{g(x)}

equal to?

Options

A $\frac{\ln 2}{2}$
B $\frac{\ln 2}{4}$
C $\ln 2$
D $2 \ln 2$

Correct answer

C. $\ln 2$

Step-by-step solution

Given that

\mathrm{g}(\mathrm{x})=\sin \mathrm{x}+\cos \mathrm{x}+1

\begin{aligned} & \int_0^{\frac{\pi}{2}} \frac{d x}{g(x)}=\int_0^{\frac{\pi}{2}} \frac{1}{\sin x+\cos x+1} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{1}{\frac{2 \tan \frac{\pi}{2}}{1+\tan ^2 \frac{\pi}{2}}+\frac{1-\tan ^2 \frac{\pi}{2}}{1+\tan ^2 \frac{\pi}{2}}+1} d x \end{aligned}

$\begin{aligned} & =\int_0^{\frac{\pi}{2}} \frac{1+\tan ^2 \frac{x}{2}}{2 \tan \frac{x}{2}+1-\tan ^2 \frac{x}{2}+1+\tan ^2 \frac{x}{2}} d x \\ & =\int_0^{\frac{\pi}{2}} \frac{\sec ^2 \frac{x}{2}}{2\left(1+\tan \frac{x}{2}\right)} d x \\ & \text{ let } 1+\tan \frac{x}{2}=t \Rightarrow \frac{1}{2} \sec ^2 \frac{x}{2} d x=d t \end{aligned}

when

\mathrm{x}=\pi / 2

then

\mathrm{t}=2

, and when

\mathrm{x}=0

then

\mathrm{t}=1=\int_1^2 \frac{1}{\mathrm{t}} \mathrm{dt}=[\ln \mathrm{t}]_1^2=\ln 2-\ln 1=\ln 2

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