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Need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.8 question 19

Answers (1)

Answer:

$\frac{1}{b-a}In\left | a\: cos^{2}\: x+b\: sin^{2}\: x \right |+C$

Hint:

$\int \! \frac{dt}{t}=log\left | t \right |+C$

Given:

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

Explanation:

Let

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

$(2a\, cos\, x(-sin\, x)+b(2\, sin\, x)(cos\, x))dx=dt$

$(b-a)[2\, cos\, x\, sin\, x]dx=dt$

$(b-a)sin\, 2x\, dx=dt$

$sin\, 2x\, dx=\frac{dt}{b-a}$

Put in (1)

$\frac{1}{b-a}\int \! \frac{dt}{t}$

$=\frac{1}{b-a}In\left | t \right |+C$

$=\frac{1}{b-a}In\left | a\: cos^{2}\: x+b\: sin^{2} \right |+C$

Posted by

Gurleen Kaur

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