Question

Integrate.

$\displaystyle \int_0^{2\pi} cos^5 x\ dx$

Answer 1

Topic :-

Definite Integration

To Integrate :-

$\displaystyle \int_{0}^{2\pi}\cos^5x\:dx$

Solution :-

$\displaystyle \int_{0}^{2\pi}\cos^5x\:dx$

$\displaystyle \int_{0}^{2\pi}\cos x.\cos^4x\:dx$

$\displaystyle \int_{0}^{2\pi}\cos x(\cos^2x)^2\:dx$

$\because \: \cos^2x=1-\sin^2x$, we can write,

$\displaystyle \int_{0}^{2\pi}\cos x(1-\sin^2x)^2\:dx$

Substitute sinx = t,

Differentiate both sides,

cosx.dx = dt

Change limits,

Lower limit :-

sin(0) = 0

Upper limit :-

sin(2π) = 0

Now, we can write,

$\displaystyle \int_{0}^{0}(1-t^2)^2\:dt$

As we know,

$\displaystyle \int_{0}^{0}f(x)\:dx=0$

So,

$\displaystyle \int_{0}^{0}(1-t^2)^2\:dt=0$

Answer :-

So, answer is Zero (0).

Additional Formulae :-

$\displaystyle \int \sin x\:dx=-\cos x+C$

$\displaystyle \int \cos x\:dx=\sin x+C$

$\displaystyle \int \tan x\:dx=ln|\sec x|+C$

$\displaystyle \int e^x\:dx=e^x+C$

Answer 2

Answer:

answer :

Topic :

Definite Integration

To Integrate :

27 cos x da

Solution :-

2T cos x dx

2л cos x. cos* x dx

2T 0 cos r (cos? a)? d

* cos x = 1 – sin x, we can write,

cos a (1 sin? x)? dx

Substitute sinx = t,

Differentiate both sides,

cosx.dx = dt

Change limits,

.

Lower limit

sin(0) = 0

Upper limit :

sin(2n) = 0

Now, we can write,

(1 – t2)² dt

As we know,

f(x) dx = 0

So,(1 -t?)2 dt = 0

Answer:

So, answer is Zero (0).

Additional Formulae :-

sin x dx cos a + C

cos x dx = sin x + C

| tan tan x dx = In sec x +C

jed e da = e" + C

hope it helps

have a wonderful day