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  • Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 14.

Need solution for RD Sharma Maths Class 12 Chapter 21 Differential Equation exercise 21.5 question 14.

Answers (1)

Answer: y=-\frac{1}{3t^3}+C

Hint: You have to integrate by applying integration of \sin^nx.

Given: \sin^4x\frac{dy}{dx}=\cos x

Solution: \frac{dy}{dx}=\frac{\cos x}{ \sin^4x}dx                                                                               …… (i)

Put t=\sin x

dt=\cos x \; dx

Integrate both sides in (i)

  \begin{aligned} &\int d y=\int \frac{d t}{t^{4}} \\ &y=\int \frac{1}{t^{4}} \\ &y=-\frac{1}{3 t^{3}}+C\\ &y=-\frac{1}{3\sin^{3}x}+C \end{aligned}

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