a-sin-3-4x-cos-8-4x-dx-b-pi-2-pi-2-x-2-e-cosx-2x-sinxdx-

Question Number 108921 by Ar Brandon last updated on 20/Aug/20

a . \int \frac{\sin^{3} 4 x}{\cos^{8} 4 x} dx

b . \int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} e^{cosx} - 2 x) sinxdx

Answered by bobhans last updated on 20/Aug/20

\frac{≊ β o β ≅}{∦∦∦∦}

(a) \int \frac{(1 - \cos^{2} 4 x) . \sin 4 x}{\cos^{8} 4 x} d x = - \frac{1}{4} \int \frac{(1 - h^{2}) d h}{h^{8}}

= - \frac{1}{4} {\int h^{- 8} d h - \int h^{- 6} d h}

= - \frac{1}{4} {- \frac{1}{7 h^{7}} + \frac{1}{5 h^{5}}} + c

= \frac{1}{28 \cos^{7} 4 x} - \frac{1}{20 \cos^{5} 4 x} + c

Commented by Ar Brandon last updated on 20/Aug/20

thanks

Answered by Ar Brandon last updated on 20/Aug/20

b . I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} e^{cosx} - 2 x) sinxdx, u = - x

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} e^{cosx} + 2 x) \sin (- x) dx

2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} e^{cosx} - 2 x - x^{2} e^{cosx} - 2 x) sinxdx

= - \int_{- \frac{π}{2}}^{\frac{π}{2}} 4 xsinxdx

I = - \int_{- \frac{π}{2}}^{\frac{π}{2}} 2 xsinxdx

Answered by Dwaipayan Shikari last updated on 20/Aug/20

\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} e^{c o s x} - 2 x) s i n x d x = \int_{- \frac{π}{2}}^{\frac{π}{2}} - (x^{2} e^{c o s x} - 2 x) s i n x d x = I

2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} - 4 x s i n x d x

- \frac{I}{4} = \int_{0}^{\frac{π}{2}} x s i n x d x

- \frac{I}{4} = - {[x c o s x]}_{0}^{\frac{π}{2}} + {[s i n x]}_{0}^{\frac{π}{2}}

I = - 4

Commented by Ar Brandon last updated on 20/Aug/20

Cool !

Commented by Ar Brandon last updated on 20/Aug/20

Hey bro, what do you think of Q108491 ? ��