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  • Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 40

Explain solution rd sharma class 12 chapter Indefinite Integrals exercise 18.2 question 40

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\begin{aligned} & a^{2} \tan x-b^{2} \cot x-(a-b) x\\ &\text { Hint: Using } \int \sec ^{2} x d x \text { and } \int \cos e c^{2} x d x\\ &\text { Given: } \int(a \tan x+b \cot x)^{2} d x\\ &\text { Solution: }\\ &I=\int(a \tan x+b \cot x)^{2} d x\\ &=\int\left(a^{2} \tan ^{2} x+b^{2} \cot ^{2} x+2 a b \tan x \cot x\right) d x \quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right]\\ &=\int\left(a^{2} \tan ^{2} x+b^{2} \cot ^{2} x+2 a b(1)\right) d x \quad\left[\begin{array}{l} \because \tan x=\frac{1}{\cot x} \\ \Rightarrow \tan x \cot x=1 \end{array}\right] \end{aligned}\begin{aligned} &=a^{2} \int \tan ^{2} x d x+b^{2} \int \cot ^{2} x d x+2 a b \int 1 d x\left[\begin{array}{l} \because \tan ^{2} \theta=\sec ^{2} \theta-1 \\ \cot ^{2} \theta=\cos e c^{2} x-1 \end{array}\right] \\ &=a^{2} \int\left(\sec ^{2} x-1\right) d x+b^{2} \int\left(\cos e c^{2} x-1\right) d x+2 a b \int 1 d x \\ &=a^{2} \tan x-a^{2} x+b^{2}(-\cot x)-b^{2} x+2 a b x \\ &I=a^{2} \tan x-b^{2} \cot x-x\left(a^{2}+b^{2}-2 a b\right) \\ &=a^{2} \tan x-b^{2} \cot x-(a-b)^{2} x \quad\left[\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right] \end{aligned}

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