Solve-cos-1-x-1-x-2-1-log-e-2-sin-2x-1-x-2-pi-dx-Evaluate-pi-2-pi-2-sin-2-xcos-2-x-cosx-sinx-dx-

Question Number 73429 by Henri Boucatchou last updated on 12/Nov/19

S o l v e : \int \frac{{[{c o s}^{- 1} x \sqrt{1 - x^{2}}]}^{- 1}}{{l o g}_{e} [2 + \frac{s i n (2 x \sqrt{1 - x^{2}})}{π}]} d x

E v a l u a t e \int_{- π / 2}^{π / 2} {s i n}^{2} {x c o s}^{2} x (c o s x + s i n x) d x

Commented by MJS last updated on 12/Nov/19

\int \sin^{2} x \cos^{2} x (\cos x + \sin x) d x =

= \int \sin^{3} x \cos^{2} x d x + \int \sin^{2} x \cos^{3} x d x =

= - \frac{\int \sin 5 x d x}{16} + \frac{\int \sin 3 x d x}{16} + \frac{\int \sin x d x}{8} -

- \frac{\int \cos 5 x d x}{16} - \frac{\int \cos 3 x d x}{16} + \frac{\int \cos x d x}{8} =

= \frac{\cos 5 x}{80} - \frac{\cos 3 x}{48} - \frac{\cos x}{8} - \frac{\sin 5 x}{80} - \frac{\sin 3 x}{48} + \frac{\sin x}{8} + C

\underset{- \frac{π}{2}}{\int^{\frac{π}{2}}} \sin^{2} x \cos^{2} x (\cos x + \sin x) d x = \frac{4}{15}

Commented by MJS last updated on 12/Nov/19

please check the first one, is it \sin or \sin^{- 1} ?

Commented by MJS last updated on 12/Nov/19

\dots anyway I cannot solve it

Answered by MJS last updated on 12/Nov/19

\int \sin^{2} x \cos^{2} x (\cos x + \sin x) d x =

= \int \sin^{3} x \cos^{2} x d x + \int \sin^{2} x \cos^{3} x d x =

[u = \cos x \to d x = - \frac{d u}{\sin x}; v = \sin x \to d x = \frac{d v}{\cos x}]

= \int u^{4} - u^{2} d u + \int v^{2} - v^{4} d v =

= \frac{u^{5}}{5} - \frac{u^{3}}{3} + \frac{v^{3}}{3} - \frac{v^{5}}{5} =

= \frac{\cos^{5} x}{5} - \frac{\cos^{3} x}{3} + \frac{\sin^{3} x}{3} - \frac{\sin^{5} x}{5} + C

\underset{- \frac{π}{2}}{\int^{\frac{π}{2}}} \sin^{2} x \cos^{2} x (\cos x + \sin x) d x = \frac{4}{15}

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