The value of e^ - \pi \over 4 + \int\limits_0^\pi \over 4 e^ - x\tan ^50xdx \over \int\limits_0^\pi \over 4 e^ - x(\tan ^49x + \tan ^51x)dx is

Definite Integration PYQ — JEE Main 2023 · 13 Apr · Shift 2 · Q32 with Solution

View written solutionFree

Correct answer: B

Let $I_n=\int_0^{\pi/4} e^{-x}\tan^n x\,dx.$ Then the given expression becomes

=\frac{e^{-\pi/4}+I_{50}}{I_{49}+I_{51}}.$$ 2. We now derive a relation among $I_{49}, I_{50}, I_{51}$. Consider $$\frac{d}{dx}(\tan^{50}x)=50\tan^{49}x\sec^2 x.

Since $\sec^2 x=1+\tan^2 x$ , we get $\frac{d}{dx}(\tan^{50}x)=50(\tan^{49}x+\tan^{51}x).$ Hence, $I_{49}+I_{51}=\frac1{50}\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x).$

Apply integration by parts in Stieltjes form:

=\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}-\int_0^{\pi/4}\tan^{50}x\,d(e^{-x}).$$ Now, $$d(e^{-x})=-e^{-x}dx,$$ so $$\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x) =\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}+\int_0^{\pi/4} e^{-x}\tan^{50}x\,dx.$$ That is, $$\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x) =\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}+I_{50}.$$ 4. Evaluate the boundary term: - At $x=\pi/4$, $\tan(\pi/4)=1$, so $$e^{-\pi/4}\tan^{50}(\pi/4)=e^{-\pi/4}.$$ - At $x=0$, $\tan 0=0$, so $$e^0\tan^{50}0=0.$$ Therefore, $$\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}=e^{-\pi/4}.$$ Thus, $$\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x)=e^{-\pi/4}+I_{50}.$$ So from step 2, $$I_{49}+I_{51}=\frac1{50}\left(e^{-\pi/4}+I_{50}\right).$$ 5. Rearranging, $$e^{-\pi/4}+I_{50}=50(I_{49}+I_{51}).$$ Therefore the required value is $$\frac{e^{-\pi/4}+I_{50}}{I_{49}+I_{51}}=50.$$ 6. Hence the correct option is: $$\boxed{50}$$ which is **Option B**.

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View written solutionFree

Correct answer: B

Let $I_n=\int_0^{\pi/4} e^{-x}\tan^n x\,dx.$ Then the given expression becomes

=\frac{e^{-\pi/4}+I_{50}}{I_{49}+I_{51}}.$$ 2. We now derive a relation among $I_{49}, I_{50}, I_{51}$. Consider $$\frac{d}{dx}(\tan^{50}x)=50\tan^{49}x\sec^2 x.

Since $\sec^2 x=1+\tan^2 x$ , we get $\frac{d}{dx}(\tan^{50}x)=50(\tan^{49}x+\tan^{51}x).$ Hence, $I_{49}+I_{51}=\frac1{50}\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x).$

Apply integration by parts in Stieltjes form:

=\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}-\int_0^{\pi/4}\tan^{50}x\,d(e^{-x}).$$ Now, $$d(e^{-x})=-e^{-x}dx,$$ so $$\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x) =\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}+\int_0^{\pi/4} e^{-x}\tan^{50}x\,dx.$$ That is, $$\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x) =\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}+I_{50}.$$ 4. Evaluate the boundary term: - At $x=\pi/4$, $\tan(\pi/4)=1$, so $$e^{-\pi/4}\tan^{50}(\pi/4)=e^{-\pi/4}.$$ - At $x=0$, $\tan 0=0$, so $$e^0\tan^{50}0=0.$$ Therefore, $$\left[e^{-x}\tan^{50}x\right]_0^{\pi/4}=e^{-\pi/4}.$$ Thus, $$\int_0^{\pi/4} e^{-x}\,d(\tan^{50}x)=e^{-\pi/4}+I_{50}.$$ So from step 2, $$I_{49}+I_{51}=\frac1{50}\left(e^{-\pi/4}+I_{50}\right).$$ 5. Rearranging, $$e^{-\pi/4}+I_{50}=50(I_{49}+I_{51}).$$ Therefore the required value is $$\frac{e^{-\pi/4}+I_{50}}{I_{49}+I_{51}}=50.$$ 6. Hence the correct option is: $$\boxed{50}$$ which is **Option B**.

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