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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefiite Integrals Excercise Fill in the Blanks Question 5

Answers (1)

Answer:   k= \frac{1}{7}

Hint: Use simple integration


\begin{aligned} &I=\int \frac{\sin ^{6} x}{\cos ^{8} x} d x \\ &I=\int \frac{\sin ^{6} x}{\cos ^{6} x \cdot \cos ^{2} x} d x \\ &I=\int \tan ^{6} x \cdot \sec ^{2} x d x \end{aligned}




\tan x = y                …(i)

\begin{aligned} &\sec ^{2} x d x=d y \\ &\Rightarrow I=\int y^{6} d y \\ &=\frac{y^{7}}{7}+c \end{aligned}


I=\frac{\tan ^{7} x}{7}+c              from equation(i)


As given

\begin{aligned} &\int \frac{\sin ^{6} x}{\cos ^{8} x} d x=k \tan ^{7} x+c \\ & \end{aligned}

\frac{\tan ^{7} x}{7}+c=k \tan ^{7} x+c

Comparing both sides

k= \frac{1}{7}

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