Question

integral 0 to pie sin^4theta d theta

Answer 1

EXPLANATION.

$\sf \implies \int\limits^\pi_0 {sin^{4} x} \, dx$

As we know that,

First we integrate the value of sin⁴ x dx, we get.

$\sf \implies \int\limits^\pi_0 sin^{2}x sin^{2}x.dx$

As we know that,

Formula of :

sin²x = (1 - cos(2x)/2)².dx

$\sf \implies \int\limits^\pi_0\bigg(\dfrac{1 - cos(2x)}{2} \bigg)^{2} .dx$

$\sf \implies \int\limits^\pi_0 \dfrac{1}{4} \bigg(1 - cos(2x) \bigg)^{2}.dx$

$\sf \implies \dfrac{1}{4} \int\limits^\pi_0 (1 + cos(2x)^{2} - 2cos(2x).dx$

As we know that,

Formula of :

⇒ cos(2∅) = 2cos²∅ - 1.

⇒ cos2∅ + 1 = 2cos²∅.

Put the value of ∅ = 2x in equation, we get.

⇒ cos2(2x) + 1 = 2cos²(2x).

⇒ cos4x + 1 = 2cos²2x.

⇒ cos4x + 1/2 = cos²2x.

Put the value of cos²2x = cos4x + 1/2, we get.

$\sf \implies \dfrac{1}{4} \int\limits^\pi_0 \bigg(1 + \dfrac{cos4x + 1}{2} - 2cos2x \bigg).dx$

$\sf \implies \dfrac{1}{4} \int\limits^\pi_0(1.dx) + \dfrac{1}{8} \int\limits^\pi_0 (cos4x + 1).dx - \dfrac{2}{4} \int\limits^\pi_0cos(2x).dx$

$\sf \implies \dfrac{1}{4} \int\limits^\pi_0(1.dx) + \dfrac{1}{8} \int\limits^\pi_0 cos(4x).dx + \dfrac{1}{8} \int\limits^\pi_0(1.dx) - \dfrac{1}{2} \int\limits^\pi_0 cos(2x).dx$

$\sf \implies \bigg[\dfrac{x}{4} + \dfrac{x}{8} + \dfrac{sin(4x)}{32} - \dfrac{sin(2x)}{4} \bigg]_0^\pi$

$\sf \implies \bigg[\dfrac{3x}{8} + \dfrac{sin(4x)}{32} - \dfrac{sin(2x)}{4} \bigg]_0^\pi$

As we know that,

First we put the upper limit then we put the lower limits in the equation, we get.

$\sf \implies \bigg[\dfrac{3\pi}{8} + \dfrac{sin(4\pi)}{32} - \dfrac{sin(2\pi)}{4} \bigg] - \bigg[\dfrac{3(0)}{8} + \dfrac{sin(4(0))}{32} - \dfrac{sin(2(0))}{4} \bigg]$

As we know that,

⇒ sin2π = 0.

⇒ sin2(2π) = 0.

$\sf \implies \int\limits^\pi_0sin^{4}x.dx = \dfrac{3\pi}{8}$

                                                                                                                                     

MORE INFORMATION.

Definition of [NEWTON-LEIBNITZ FORM].

Let ''f'' if function of x defined on [a, b] and d[f(x)]/dx = Ф(x) and a and b are two values independent of variable x, then for all values of x in domain of ''f'' then,

$\sf \implies \int\limits^b_a {\phi(x)} \, dx = [f(x)]_a^b = f(b) - f(a).$

Answer 2

IDENTITIES USED :-

$\longmapsto \boxed{ \green{ \bf \: \: \int\limits _0 ^{ a } f(x) \: dx = \: \int\limits _0 ^{ a }f(a - x)dx }}$

$\boxed{ \green{ \bf \: \: \int\limits _0 ^{ 2a } f(x) \: dx = 2\: \int\limits _0 ^{ a }f(x)dx \: when \: f(2a - x) = f(x)}}$

CALCULATION

$\tt \implies \: Let \: I \: = \: \int\limits^\pi_0 {sin^{4} x} \, dx - - - (1)$

$\longmapsto \rm \: Let \: f(x) = {sin}^{4} x$

So,

$\longmapsto \rm \: f(\pi \: - x) = {sin}^{4} (\pi \: - x)$

$\longmapsto \rm \: f(\pi \: - x) = {sin}^{4} x$

$\rm :\implies\:f(\pi \: - x) = f(x)$

So,

$\tt\: I \: = \: 2 \int\limits _0 ^{ \frac{\pi}{2} } {sin^{4} x} \, dx - - (2)$

Now,

$\longmapsto \boxed{ \green{ \bf \:change \: x \: \to \: \dfrac{\pi}{2} - x}}$

$\tt\: I \: = \: 2 \int\limits _0 ^{ \frac{\pi}{2} } {sin^{4} \bigg(\dfrac{\pi}{2} - x \bigg)} \, dx$

$\tt\: I \: = \: 2 \int\limits _0 ^{ \frac{\pi}{2} } {cos^{4} x} \, dx - - - (3)$

Now,

Adding equation (2) and equation (3), we get

$\rm :\implies\:\tt\: 2I \: = \: 2 \int\limits _0 ^{ \frac{\pi}{2} } ({sin^{4} x} + {cos^{4} x})\, dx$

$\rm :\implies\:\tt\: I \: = \: \int\limits _0 ^{ \frac{\pi}{2} }( ({sin^{2} x})^{2} + ({cos^{2} x}) ^{2}) \, dx$

$\rm :\implies\:\tt\: I \: = \: \int\limits _0 ^{ \frac{\pi}{2} } {(( {sin}^{2}x + {cos}^{2}x) }^{2} - 2 {sin}^{2} x {cos}^{2} x) \: dx$

$\rm :\implies\:I = \: \int\limits _0 ^{ \frac{\pi}{2} } (1 - 2 {sin}^{2} x {cos}^{2} x)d$

$\rm :\implies\: I = \: \int\limits _0 ^{ \frac{\pi}{2} } 1dx - \dfrac{1}{2} \: \int\limits _0 ^{ \frac{\pi}{2} } 4 {sin}^{2} x {cos}^{2} x \: dx$

$\rm :\implies\:I = \:(x) _0 ^{ \frac{\pi}{2} } - \dfrac{1}{2} \: \int\limits _0 ^{ \frac{\pi}{2} } {sin}^{2} 2x \: dx$

$\rm :\implies\:I = \dfrac{\pi}{2} - \dfrac{1}{2} \: \int\limits _0 ^{ \frac{\pi}{2} } \dfrac{1 - cos4x}{2} dx$

$\rm :\implies\:I = \dfrac{\pi}{2} - \dfrac{1}{4} \: \int\limits _0 ^{ \frac{\pi}{2} }(1 - cos4x)dx$

$\rm :\implies\:I = \dfrac{\pi}{2} - \dfrac{1}{4} \: \bigg(x - \dfrac{sin4x}{4} \bigg) _0 ^{ \frac{\pi}{2} }$

$\rm :\implies\:I = \dfrac{\pi}{2} - \dfrac{1}{4} (\dfrac{\pi}{2} )$

$\rm :\implies\:I = \dfrac{\pi}{2} - (\dfrac{\pi}{8} )$

$\rm :\implies\:I = \dfrac{4\pi \: - 3\pi}{8}$

$\rm :\implies\:I = \dfrac{3\pi}{8}$