Evaluate \int_0^{\frac{\pi}{2}} \frac{1}{a \sin^2 x+b \cos^2 x}dx

Ben Shaver

Ben Shaver

Answered question

2022-01-16

Evaluate 0π21asin2x+bcos2xdx

Answer & Explanation

Anzante2m

Anzante2m

Beginner2022-01-17Added 34 answers

Hint
1asin2x+bcos2x=sec2xatan2x+b
Let atanx=btany
x=0y=0  and  x=π2y=?
Kayla Kline

Kayla Kline

Beginner2022-01-18Added 37 answers

By the substitution u=cotg(x), the integral becomes
+0dua+bu2
=1a0+du1+bau2
now, put v=uba
it gives
1a0+abdv1+v2
=1ab[arctan(v)]0+=π2ab

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