# Get Answers to all your Questions

#### Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 44

\begin{aligned} &=-3 \cos x-4 \sin x+6 \tan x+7 \cot x+c \end{aligned}

Hint: You must know about integration of all trigonometry function

\begin{aligned} &\text { Given: } \int\left\{3 \sin x-4 \cos x+\frac{5}{\cos ^{2} x}-\frac{6}{\sin ^{2} x}+\tan ^{2} x-\cot ^{2} x\right\} d x\\ &\text { Solution: } \int 3 \sin x d x-\int 4 \cos x d x+5 \int \frac{1}{\cos ^{2} x} d x-6 \int \frac{1}{\sin ^{2} x} d x+\int \tan ^{2} x d x-\int \cot ^{2} x d x\\ &=\int 3 \sin x d x-\int 4 \cos x d x+5 \int \sec ^{2} x d x-6 \int \operatorname{cosec}^{2} x d x+\int\left(\sec ^{2} x-1\right) d x-\int\left(\operatorname{cosec}^{2} x-1\right) d x\\ &\left[\because \sec \theta=\frac{1}{\cos \theta} ; \operatorname{cosec} \theta=\frac{1}{\sin \theta} ; \tan ^{2} \theta=\sec ^{2} \theta-1 ; \cot ^{2} \theta=\operatorname{cosec}^{2} \theta-1\right]\\ &=3(-\cos x)-4 \sin x+5 \tan x-6(-\cot x)+\tan x-x-(-\cot x)+x+c\\ &=-3 \cos x-4 \sin x+6 \tan x+7 \cot x+c \end{aligned}