#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 19 textbook solution.

Answer : $y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C$

Hint : To solve this we convert  $\cot x \text { to } \frac{\cos x}{\sin x}$  formula.

Give : $\frac{d y}{d x}+y \tan x=x^{2} \cos ^{2} x$

Solution : $\frac{dy}{dx}+Py=Q$

\begin{aligned} &P=\tan x, Q=x^{2} \cos ^{2} x \\ \end{aligned}

\begin{aligned} &I f=e^{\int P d x} \\ \end{aligned}

\begin{aligned} &=e^{\int \tan x d x} \\ \end{aligned}

\begin{aligned} &=e^{\log \sec x} \\ \end{aligned}

\begin{aligned} &=\sec x \end{aligned}

\begin{aligned} &=y I f=\int Q I f d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(\sec x) d x+C \\ &=y \sec x=\int x^{2} \cos ^{2} x(1 / \cos x) d x+C \\ &=y \sec x=\int x^{2} \cos x d x+C \end{aligned}

$=y \sec x=x^{2} \int \cos x d x-\int(2 x \cos x d x) d x+C$

Using integration by parts

\begin{aligned} &=y \sec x=x^{2} \sin x-2 \int x^{2} \sin x d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x-2 \int x \sin x d x-\int(\sin x d x) d x+C \\ &=y \sec x=x^{2} \sin x+2 x \cos x-2 \sin x+C \end{aligned}