First slide

Methods of integration

Question

$\begin{array}{l} If \int \frac{\cos x d x}{\sin^{3} x {(1 + \sin^{6} x)}^{2 / 3}} = f (x) {(1 + \sin^{6} x)}^{1 / λ} + c where c is a constant of integration, \\ then λ f (\frac{π}{3}) is equal to: \end{array}$

Very difficult

A

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

-2
C

2
D

$\frac{9}{8}$

Solution

$\begin{array}{l} L e t \sin x = t \\ \Rightarrow \cos x d x = d t \\ N o w I = \int \frac{d t}{t^{3} {(1 + t^{6})}^{\frac{2}{3}}} = \int \frac{d t}{t^{7} {(1 + \frac{1}{t^{6}})}^{\frac{2}{3}}} \\ Put 1 + \frac{1}{t^{6}} = r^{3} \Rightarrow \frac{d t}{t^{7}} = \frac{- 1}{2} r^{2} d r \\ \Rightarrow - \frac{1}{2} \int \frac{r^{2} d r}{r^{2}} = - \frac{1}{2} r + c = - \frac{1}{2} {(\frac{\sin^{6} x + 1}{\sin^{6} x})}^{\frac{1}{3}} + c = - \frac{1}{2 \sin^{2} x} {(1 + \sin^{6} x)}^{\frac{1}{3}} + c \\ \Rightarrow f (x) = - \frac{1}{2} {cosec}^{2} x and λ = 3 \\ ∴ λ f (\frac{π}{3}) = - 2 \end{array}$

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