I want to evaluate \(\displaystyle\int{\frac{{{\cos{\theta}}{d}\theta}}{{\sqrt{{{2}-{9}{{\sin}^{{2}}\theta}}}}}}\)

Emilia Hoffman

Emilia Hoffman

Answered question

2022-04-07

I want to evaluate
cosθdθ29sin2θ

Answer & Explanation

tutaonana223a

tutaonana223a

Beginner2022-04-08Added 15 answers

Using the substitution u=23sinθdudθ=23cosθ. Our integral can be written as
cosθ29sin2dθdudu=cosθ29sin2θ
So
32129sin2θdu=3212(81u22)du
And hence we get
32(81u22)du=13arcsin(9u2)+C
Back substituting yields
13arcsin(3sinθ2)+C

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