Integration by Substitution?

Question

Integration by Substitution?

Show that:
$\int_{0}^{\frac{π}{2}} {sin}^{2} (x) d x = \int_{0}^{\frac{π}{2}} {cos}^{2} (x) d x$ $int_0^(pi/2)sin^2(x)dx=int_0^(pi/2)cos^2(x)dx$

3 Answers

Impact of this question

1660 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-25T12:42:16

$\int_{0}^{\frac{π}{2}} {cos}^{2} (x) d x = \int_{0}^{\frac{π}{2}} {sin}^{2} (x) d x$ $int_0^(pi/2) cos^2(x)dx = int_0^(pi/2) sin^2(x)dx$

$\int_{0}^{\frac{π}{2}} {cos}^{2} (x) d x - \int_{0}^{\frac{π}{2}} {sin}^{2} (x) d x = 0$ $int_0^(pi/2) cos^2(x)dx - int_0^(pi/2) sin^2(x)dx = 0$

using the linearity of the integral:

$\int_{0}^{\frac{π}{2}} ({cos}^{2} (x) d x - {sin}^{2} (x)) d x = 0$ $int_0^(pi/2) (cos^2(x)dx - sin^2(x))dx = 0$

and the trigonometric identity: $cos (2 α) = {cos}^{2} α - {sin}^{2} α$ $cos(2alpha) = cos^2alpha -sin^2alpha$

$\int_{0}^{\frac{π}{2}} cos (2 x) d x = 0$ $int_0^(pi/2) cos(2x)dx = 0$

In fact:

$\int_{0}^{\frac{π}{2}} cos (2 x) d x = \frac{1}{2} {[sin (2 x)]}_{0}^{\frac{π}{2}} = 0$ $int_0^(pi/2) cos(2x)dx = 1/2 [sin(2x)]_0^(pi/2) = 0$

which proves the point.

Answer 2 · 2017-07-25T12:46:18

$\int_{0}^{\frac{π}{2}} {sin}^{2} x d x = \int_{0}^{\frac{π}{2}} {cos}^{2} x d x \Leftrightarrow$ $int_0^(pi/2)sin^2xdx=int_0^(pi/2)cos^2xdx iff$

$\int_{0}^{\frac{π}{2}} {cos}^{2} x d x - \int_{0}^{\frac{π}{2}} {sin}^{2} x d x = 0 \Leftrightarrow$ $int_0^(pi/2)cos^2xdx-int_0^(pi/2)sin^2xdx=0 iff$

$\int_{0}^{\frac{π}{2}} ({cos}^{2} x - {sin}^{2} x) d x = 0 \Leftrightarrow$ $int_0^(pi/2)(cos^2x-sin^2x)dx=0iff$

$\int_{0}^{\frac{π}{2}} cos 2 x d x = 0 \Leftrightarrow$ $int_0^(pi/2)cos2xdx=0 iff$

Let's substitude $u = 2 x \Rightarrow d u = 2 d x \Rightarrow d x = \frac{d u}{2}$ $u=2x =>du=2dx=>dx=(du)/2$ :

$\int_{0}^{π} cos u \frac{d u}{2} = 0 \Leftrightarrow$ $int_0^picosu(du)/2=0iff$

$\frac{1}{2} {[sin u]}_{0}^{π} = 0 \Leftrightarrow (sin π - sin 0) = 0 \Leftrightarrow$ $1/2[sinu]_0^pi=0iff(sinpi-sin0)=0iff$

$0 = 0$ $0=0$ which is true, so the first statement is true.

Answer 3 · 2017-07-25T12:59:57

Kindly, refer to the Explanation.

Explanation:

Using the well-known Result : $\int_{0}^{a} f (x) d x = \int_{0}^{a} f (a - x) d x,$ $int_0^af(x)dx=int_0^af(a-x)dx,$

we have, $\int_{0}^{\frac{π}{2}} {sin}^{2} x d x = \int_{0}^{\frac{π}{2}} {sin}^{2} (\frac{π}{2} - x) d x,$ $int_0^(pi/2)sin^2xdx=int_0^(pi/2)sin^2(pi/2-x)dx,$

$= \int_{0}^{\frac{π}{2}} {cos}^{2} x d x .$ $=int_0^(pi/2)cos^2xdx.$

Hence, the Proof.

Science

Math

Humanities

... and beyond

Integration by Substitution?

Show that:
$\int_{0}^{\frac{π}{2}} {sin}^{2} (x) d x = \int_{0}^{\frac{π}{2}} {cos}^{2} (x) d x$ $int_0^(pi/2)sin^2(x)dx=int_0^(pi/2)cos^2(x)dx$

3 Answers

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Integration by Substitution?

Show that: ∫π20sin2(x)dx=∫π20cos2(x)dxint_0^(pi/2)sin^2(x)dx=int_0^(pi/2)cos^2(x)dx

3 Answers

Explanation:

Related questions

Impact of this question

Show that:
$\int_{0}^{\frac{π}{2}} {sin}^{2} (x) d x = \int_{0}^{\frac{π}{2}} {cos}^{2} (x) d x$ $int_0^(pi/2)sin^2(x)dx=int_0^(pi/2)cos^2(x)dx$