Q46 / 104 · NTA Abhyas (JEE Main) 2020 · Mathematics · Indefinite Integration

The integral $I = \int \sec^{3} x \tan^{3} x d x$ is equal to (where, $C$ is the constant of integration)

A
$\sec^{5} x - \sec^{3} x + C$
B
$\frac{\sec^{5} x}{5} - \sec^{3} x + C$
Correct answer
C
$\frac{\sec^{5} x}{5} - \frac{\sec^{3} x}{3} + C$
D
$\frac{\sec^{5} x}{5} - \tan^{- 1} x + C$

Solution

Let

\sec x = t \Rightarrow \sec x \tan x d x = d t

∴ I = \int \sec^{2} x \tan^{2} x (\sec x \tan x d x)

= \int t^{2} (t^{2} - 1) d t

= \int (t^{4} - t^{2}) d t

= \frac{t^{5}}{5} - \frac{t^{3}}{3} + C

I = \frac{\sec^{5} x}{5} - \frac{\sec^{3} x}{3} + C

Hence, the correct answer is option: 'C'.

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