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Need solution for RD Sharma Maths Class 12 Chapter 18 Indefinite Integrals Excercise Very Short Answers Question 25

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Answer:\frac{1}{a^{2}-b^{2}}\log_{e}\left | a^{2}\sin^{2}x+b^{2}\cos ^{2}x \right |+c

Hints: You must know about the integral rule of trigonometric functions

Given: \int \frac{\sin2x}{a^{2}\sin ^{2}x+b^{2}\cos^{2}x}dx

Solution: \int \frac{\sin2x}{a^{2}\sin ^{2}x+b^{2}\cos^{2}x}dx


a^{2} \sin ^{2} x+b^{2} \cos ^{2} x=t \\ a^{2} 2 \sin x \cdot \cos x d x+b^{2} 2 \cos x(-\sin x)=d t \\                        \\\quad \quad \quad \quad \quad \begin{bmatrix} \frac{d}{d x} \sin x=\cos x \\\\ \frac{d}{d x} \cos x=-\sin x \end{bmatrix}\&\\\left[\frac{d}{d x} x^{n}=n x^{n-1}\right]

\begin{aligned} &\sin 2 x \cdot d x\left(a^{2}-b^{2}\right)=d t \\ &\sin 2 x d x=\frac{d t}{\left(a^{2}-b^{2}\right)} \\ &{[2 \sin x \cos x=\sin 2 x]} \\ &\int \frac{d t}{\left(a^{2}-b^{2}\right) t} \\ &\frac{1}{a^{2}-b^{2}} \int \frac{d t}{t} \\ &\int \frac{1}{x} d x=\log |x|+c \\ &=\frac{1}{a^{2}-b^{2}} \log _{e}|t|+c \\ &=\frac{1}{a^{2}-b^{2}} \log _{e}\left|a^{2} \sin ^{2} x+b^{2} \cos ^{2} x\right|+c \end{aligned}

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