Question

Integre: (cos x)^4

O ^4 é elevado a 4.

Answer 1

Resposta:

∫ cos⁴ x dx  

∫ [cos²x]² x dx  

***cos(x+x)=cos²(x)-sen²(x)=cos²(x)+cos²(x)-1

***cos(2x)=2cos²(x)-1

***cos²(x)= (1/2) * [1 +cos(2x)]

∫ [(1/4) * [1 +cos(2x)]]² x dx  

(1/4) * ∫  1+ 2*cos(2x) + cos²(2x)  dx

(1/4) * ∫ dx = x/4 + c₁

(1/2) * ∫ cos(2x)  dx = (1/4) * sen(2x) + c₂

(1/4)* ∫  cos²(2x)  dx

***cos²(2x) =(1/2) * [1+cos(4x)]

(1/8)* ∫  (1/2) * [1+cos(4x)]  dx

(1/8)* ∫  1+cos(4x)  dx  =  x/8 + (1/32) sin(4x) + c₃

c₁ + c₂ +c₃ =c

∫ cos⁴ x dx   = x/4 +(1/4) * sen(2x) +x/8 + (1/32) sin(4x) + c

∫ cos⁴ x dx   = 3x/8 +(1/4) * sen(2x) + (1/32) sin(4x) + c

Answer 2

$\tt \displaystyle \int \left(cos\:x\right)^4\\\\\\\tt =\int cos ^4\left(x\right)dx\\\\\\\tt =\int cos ^3\left(x\right)cos \left(x\right)dx\\\\\\\tt =cos ^3\left(x\right)sin \left(x\right)-\int \:-3cos ^2\left(x\right)sin ^2\left(x\right)dx\\\\\\\tt =cos ^3\left(x\right)sin \left(x\right)-\left(-\frac{3}{8}\left(x-\frac{1}{4}sin \left(4x\right)\right)\right)\\\\\\\tt =cos ^3\left(x\right)sin \left(x\right)+\frac{3}{8}\left(x-\frac{1}{4}sin \left(4x\right)\right)$

$\to \boxed{\tt =cos ^3\left(x\right)sin \left(x\right)+\frac{3}{8}\left(x-\frac{1}{4}sin \left(4x\right)\right)+C}$