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Provide solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 38

Answers (1)

Answer:

$\frac{1}{b^{2}}log(a^{2}+b^{2}sin^{2}x)+C$

Hint:

Put denominator = t

Given:

$\int \! \frac{sin\, 2x}{a^{2}+b^{2}sin^{2}x}dx$ ......(1)

Explanation:

Let

$a^{2}+b^{2}sin^{2}x=t$

$b^{2}(2sin\, xcos\, x)dx=dt$

$b^{2}sin\, 2xdx=dt$

$sin\, 2xdx=\frac{dt}{b^{2}}$

Put in (1) we get

$\int \! \frac{dt}{b^{2}(t)}$

$=\frac{1}{b^{2}}\int \! \frac{dt}{t}$

$=\frac{1}{b^{2}}log\left | t \right |+C$

$=\frac{1}{b^{2}}log(a^{2}+b^{2}sin^{2}x)+C$

Posted by

Gurleen Kaur

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