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Problem ∫ sin x d x a 2 + b 2 cos 2 x = \int \frac{\sin x\,dx}{a^2 + b^2\cos^2 x} = ∫ a 2 + b 2 c o s 2 x s i n x d x = log ( a 2 + b 2 cos 2 x ) + c \log(a^2 + b^2\cos^2 x) + c log ( a 2 + b 2 cos 2 x ) + c 1 a b tan − 1 ( a cos x b ) + c \frac{1}{ab}\tan^{-1}\left(\frac{a\cos x}{b}\right) + c ab 1 tan − 1 ( b a c o s x ) + c 1 a b cot − 1 ( b cos x a ) + c \frac{1}{ab}\cot^{-1}\left(\frac{b\cos x}{a}\right) + c ab 1 cot − 1 ( a b c o s x ) + c 1 a b cot − 1 ( a cos x b ) + c \frac{1}{ab}\cot^{-1}\left(\frac{a\cos x}{b}\right) + c ab 1 cot − 1 ( b a c o s x ) + c log ( a 2 + b 2 cos 2 x ) + c \log(a^2 + b^2\cos^2 x) + c log ( a 2 + b 2 cos 2 x ) + c 1 a b tan − 1 ( a cos x b ) + c \frac{1}{ab}\tan^{-1}\left(\frac{a\cos x}{b}\right) + c ab 1 tan − 1 ( b a c o s x ) + c 1 a b cot − 1 ( b cos x a ) + c \frac{1}{ab}\cot^{-1}\left(\frac{b\cos x}{a}\right) + c ab 1 cot − 1 ( a b c o s x ) + c 1 a b cot − 1 ( a cos x b ) + c \frac{1}{ab}\cot^{-1}\left(\frac{a\cos x}{b}\right) + c ab 1 cot − 1 ( b a c o s x ) + c Browse the problem library