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NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

Let Ι = ∫ cos^ 3 x 1 + sin^ 2 x dx, then Ι is equal to (where c is...

Let

Ι = ∫ \dfrac{cos^{3}x}{1 + sin^{2}x}dx,

then

Ι

is equal to (where

c

is the constant of integration)

Options

A $2tan^{- 1}\left(x\right) + sin x + c$
B $2tan^{- 1}\left(sin x\right) - sin x + c$
C $2tan^{- 1}\left(x\right) - x + c$
D $2tan^{- 1}\left(sin x\right) + sin x + c$

Correct answer

B. $2tan^{- 1}\left(sin x\right) - sin x + c$

Step-by-step solution

Let,

sin x = t ⇒ cos xdx = dt

Ι = ∫ \dfrac{cos^{2}x}{1 + t^{2}}dt = ∫ \dfrac{1 - sin^{2}x}{1 + t^{2}}dt = ∫ \dfrac{1 - t^{2}}{1 + t^{2}}dt

or

Ι = ∫ \dfrac{2 - \left(1 + t^{2}\right)}{1 + t^{2}}dt

= ∫ \dfrac{2dt}{1 + t^{2}} -∫ dt

= 2tan^{- 1}t - t + c

= 2tan^{- 1}\left(sin x\right) - sin x + c

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