The value of \int\limits_0^1 8\log ( 1 + x ) \over 1 + x^2 dx is

Definite Integration PYQ — JEE Main 2011 · Shift 0 · Q40 with Solution

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Correct answer: D

We need to evaluate this definite integral.

Use the substitution $x=\tan\theta.$ Then $dx=\sec^2\theta\,d\theta, \qquad 1+x^2=1+\tan^2\theta=\sec^2\theta.$ So, $\frac{dx}{1+x^2}=d\theta.$

Also, when $x=0$ , $\theta=0$ , and when $x=1$ , $\theta=\frac{\pi}{4}$ .

Hence, $I=8\int_0^{\pi/4} \log(1+\tan\theta)\,d\theta.$

Simplify $1+\tan\theta$ : $1+\tan\theta=\frac{\sin\theta+\cos\theta}{\cos\theta}.$ Thus, $\log(1+\tan\theta)=\log(\sin\theta+\cos\theta)-\log(\cos\theta).$ So, $I=8\int_0^{\pi/4}\log(\sin\theta+\cos\theta)\,d\theta-8\int_0^{\pi/4}\log(\cos\theta)\,d\theta.$
Use the identity $\sin\theta+\cos\theta=\sqrt{2}\cos\left(\theta-\frac{\pi}{4}\right).$ Therefore, $\log(\sin\theta+\cos\theta)=\frac12\log 2+\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right).$ Hence,

=\int_0^{\pi/4}\left(\frac12\log 2+\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right)\right)d\theta.$$ So, $$=\frac{\pi}{8}\log 2+\int_0^{\pi/4}\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right)d\theta.$$ 5. In the last integral, put $$u=\frac{\pi}{4}-\theta \quad \Rightarrow \quad d\theta=-du.$$ As $\theta:0\to \frac{\pi}{4}$, we get $u:\frac{\pi}{4}\to 0$. Thus, $$\int_0^{\pi/4}\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right)d\theta =\int_0^{\pi/4}\log(\cos u)\,du.$$ Therefore, $$\int_0^{\pi/4}\log(\sin\theta+\cos\theta)\,d\theta =\frac{\pi}{8}\log 2+\int_0^{\pi/4}\log(\cos u)\,du.$$ 6. Substitute this back into $I$: $$I=8\left(\frac{\pi}{8}\log 2+\int_0^{\pi/4}\log(\cos u)\,du\right)-8\int_0^{\pi/4}\log(\cos\theta)\,d\theta.$$ The two cosine integrals cancel, giving $$I=8\cdot \frac{\pi}{8}\log 2=\pi\log 2.$$ 7. Therefore, $$\boxed{\int_0^1 \frac{8\log(1+x)}{1+x^2}\,dx=\pi\log 2.}$$ 8. Comparing with the options: - A: $\frac{\pi}{8}\log 2$ - B: $\frac{\pi}{2}\log 2$ - C: $\log 2$ - D: $\pi\log 2$ So the correct option is $$\boxed{\text{D}}.$$ 9. Comparison with stored correct answer: Stored correct answer = D. Our derived answer = D. They agree.

View written solutionFree

Correct answer: D

We need to evaluate this definite integral.

Use the substitution $x=\tan\theta.$ Then $dx=\sec^2\theta\,d\theta, \qquad 1+x^2=1+\tan^2\theta=\sec^2\theta.$ So, $\frac{dx}{1+x^2}=d\theta.$

Also, when $x=0$ , $\theta=0$ , and when $x=1$ , $\theta=\frac{\pi}{4}$ .

Hence, $I=8\int_0^{\pi/4} \log(1+\tan\theta)\,d\theta.$

Simplify $1+\tan\theta$ : $1+\tan\theta=\frac{\sin\theta+\cos\theta}{\cos\theta}.$ Thus, $\log(1+\tan\theta)=\log(\sin\theta+\cos\theta)-\log(\cos\theta).$ So, $I=8\int_0^{\pi/4}\log(\sin\theta+\cos\theta)\,d\theta-8\int_0^{\pi/4}\log(\cos\theta)\,d\theta.$
Use the identity $\sin\theta+\cos\theta=\sqrt{2}\cos\left(\theta-\frac{\pi}{4}\right).$ Therefore, $\log(\sin\theta+\cos\theta)=\frac12\log 2+\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right).$ Hence,

=\int_0^{\pi/4}\left(\frac12\log 2+\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right)\right)d\theta.$$ So, $$=\frac{\pi}{8}\log 2+\int_0^{\pi/4}\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right)d\theta.$$ 5. In the last integral, put $$u=\frac{\pi}{4}-\theta \quad \Rightarrow \quad d\theta=-du.$$ As $\theta:0\to \frac{\pi}{4}$, we get $u:\frac{\pi}{4}\to 0$. Thus, $$\int_0^{\pi/4}\log\left(\cos\left(\theta-\frac{\pi}{4}\right)\right)d\theta =\int_0^{\pi/4}\log(\cos u)\,du.$$ Therefore, $$\int_0^{\pi/4}\log(\sin\theta+\cos\theta)\,d\theta =\frac{\pi}{8}\log 2+\int_0^{\pi/4}\log(\cos u)\,du.$$ 6. Substitute this back into $I$: $$I=8\left(\frac{\pi}{8}\log 2+\int_0^{\pi/4}\log(\cos u)\,du\right)-8\int_0^{\pi/4}\log(\cos\theta)\,d\theta.$$ The two cosine integrals cancel, giving $$I=8\cdot \frac{\pi}{8}\log 2=\pi\log 2.$$ 7. Therefore, $$\boxed{\int_0^1 \frac{8\log(1+x)}{1+x^2}\,dx=\pi\log 2.}$$ 8. Comparing with the options: - A: $\frac{\pi}{8}\log 2$ - B: $\frac{\pi}{2}\log 2$ - C: $\log 2$ - D: $\pi\log 2$ So the correct option is $$\boxed{\text{D}}.$$ 9. Comparison with stored correct answer: Stored correct answer = D. Our derived answer = D. They agree.