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Indefinite Integrals

Section: Mathematics

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Question : 55 of 90

Marks: +1, -0

If

\frac{\int \cos x \, dx}{\sin^3 x (1+\sin^6 x)^{2/3}}

= f(x) (1+\sin^6 x)^{1/\lambda} + c

where c is a constant o f integration, then

\lambda f(\pi/3)

[8 Jan 2020 Shift 1]

-2
$-\frac{9}{8}$
2
$\frac{9}{8}$

Solution:

\frac{\int \cos x \, dx}{\sin^3 x (1+\sin^6 x)^{2/3}}

= \frac{-6}{-6} \int \frac{\cos x \, dx}{\sin^7 x \left( \frac{1}{\sin^6 x} + 1 \right)^{2/3}}

= -\frac{1}{6} \times 3 \left( \frac{1}{\sin^6 x} + 1 \right)^{1/3} + c

= -\frac{1}{2} \frac{(1+\sin^6 x)^{1/3}}{\sin^x} + c

Hence,

\lambda = 3

and

f(x) = -\frac{1}{2 \sin^2 x}

So,

\lambda f(\pi/3) = -2

REMARK : Technically, this question should be marked as bonus. Because f(x) and λ cannot be found uniquely.

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