The integral I = ∫ ^ 3 x ^ 3 xdx is equal to (where, C is the constant of integration)

Correct answer: C. ^ 5 x 5 - ^ 3 x 3 + C Let sec x = t ⇒ sec xtan xdx = dt ∴ I = ∫ ^ 2 x tan ^ 2 x( x xdx) = ∫ t ^ 2 ( t ^ 2 - 1)dt = ∫ ( t ^ 4 - t ^ 2 )dt = t ^ 5 5 - t ^ 3 3 + C I = ^ 5 x 5 - ^ 3 x 3 + C

NTA Abhyas 2020

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

NTA Abhyas 2020 · Mathematics · Indefinite Integration

The integral I = ∫ ^ 3 x ^ 3 xdx is equal to (where, C is the...

The integral

I = ∫ \sec^{3} x\tan^{3} xdx

is equal to (where,

C

is the constant of integration)

Options

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

Correct answer

C. $\dfrac{\sec^{5} x}{5} - \dfrac{\sec^{3} x}{3} + C$

Step-by-step solution

Let

sec x = t ⇒ sec xtan xdx = dt

∴ I = ∫ \sec^{2} x{tan}^{2} x\left(\sec x \tan xdx\right)

= ∫ {t}^{2}\left({t}^{2} - 1\right)dt

= ∫ \left({t}^{4} - {t}^{2}\right)dt

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

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

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