If the integral I = ∫ tan x 5 + 7 ^ 2 x dx = kln |f(x)| + C (where C is the integration constant) and f(0) = 5 , then the value of f((π)/(4)) is equal to

Correct answer: 6 I = ∫ sin xcos x 5 ^ 2 x + 7 ^ 2 x dx Let ^ 2 x = t ⇒ sin xcos xdx = (dt)/(2) I = (1)/(2)∫ (dt)/(5(1 - t) + 7t) = (1)/(2)∫ (dt)/(5 + 2t) = (1)/(4) |5 + 2t| + C = (1)/(4) |5 + 2 ^ 2 x| + C ∴ f(x) = 5 + 2 ^ 2 x ⇒ f((π)/(4)) = 6

NTA Abhyas 2020

Exam: NTA Abhyas
Year: 2020
Subject: Mathematics
Chapter: Indefinite Integration

NTA Abhyas 2020 · Mathematics · Indefinite Integration

If the integral I = ∫ tan x 5 + 7 ^ 2 x dx = kln |f(x)| + C (where...

If the integral

I = ∫ \dfrac{tan x}{5 + 7\tan^{2} x}dx

= kln \left|f\left(x\right)\right| + C

(where

C

is the integration constant)

and f\left(0\right) = 5

, then the value of

f\left(\dfrac{\pi}{4}\right)

is equal to

Correct answer

Step-by-step solution

I = ∫ \dfrac{sin xcos x}{5\cos^{2} x + 7\sin^{2} x}dx

Let

\sin^{2} x = t

⇒ sin xcos xdx = \dfrac{dt}{2}

I = \dfrac{1}{2}∫ \dfrac{dt}{5\left(1 - t\right) + 7t}

= \dfrac{1}{2}∫ \dfrac{dt}{5 + 2t}

= \dfrac{1}{4}\ln \left|5 + 2t\right| + C

= \dfrac{1}{4}\ln \left|5 + 2\sin^{2} x\right| + C

∴ f\left(x\right) = 5 + 2\sin^{2} x

⇒ f\left(\dfrac{\pi}{4}\right) = 6

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