Answer to Question #136207 in Calculus for qwerty

Question

\int \:sin^2xdx

Answer 1

Solution

$\int \:sin^2x dx$ Here, we apply reduction formula which is illustrated below;

$\int \:sin^nxdx =$ $\int \:sin^{n-1}(x).sin (x)dx$ (When spread by reduction formula)

Now, $\int \:sin^{n-1}(x).sin (x)dx$ can be solved by Integration by parts, i.e.

$\int udv = uv - \int vdu$ Where;

$u = sin^{n-1}(x)$

$dv = sin (x)dx$

$v = - cos (x)$

$du = (n-1)sin^{n-2} (x)cos (x)dx$

Therefore,

$\int \:sin^{n-1}(x).sin (x)dx = - sin^{n-1}(x) cos (x) +(n-1) \int sin^{n-2} (x)cos^2 (x) dx$

But $cos ^2(x) = 1-sin^2(x)$

$\therefore \int \:sin^{n-1}(x).sin (x)dx = - sin^{n-1}(x) cos (x) +(n-1) \int sin^{n-2} (x)(1-sin^2(x))dx$

$\implies \int \:sin^{n-1}(x).sin (x)dx = - sin^{n-1}(x) cos (x) +(n-1) \int (sin^{n-2} (x)-sin^n(x)) dx$

$\implies \int \:sin^{n-1}(x).sin(x)dx = - sin^{n-1}(x) cos (x) +(n-1) [\int sin^{n-2} (x)dx- \int sin^n(x) dx]$

Bringing back the original integral;

$\implies \int \:sin^n(x)dx = - sin^{n-1}(x) cos (x) +(n-1) \int sin^{n-2} (x)dx- (n-1)\int sin^n(x) dx$

$\therefore 1\int \:sin^n(x)dx + (n-1)\int sin^n(x) dx= - sin^{n-1}(x) cos (x) +(n-1) \int (sin^{n-2} (x)dx$

Which is; $n\int sin^n(x)dx = - sin^{n-1}(x) cos (x) +(n-1) \int (sin^{n-2} (x)dx$

${n\int sin^n(x)dx \above{2pt} n} = {- sin^{n-1}(x) cos (x) \above{2pt} n} + {(n-1) \int (sin^{n-2} (x)dx \above{2pt} n}$

$\implies \int sin^n(x)dx = -{1 \above{2pt} n} sin^{n-1}(x) cos (x) + {n-1 \above{2pt} n} \int (sin^{n-2} (x) dx$

We can rearrange such that;

$\implies \int sin^n(x)dx ={n-1 \above{2pt} n} \int sin^{n-2} (x) dx -{1 \above{2pt} n} sin^{n-1}(x) cos (x)$

Now, with $n =2 ,$ we have;

$\int sin^2(x)dx ={2-1 \above{2pt} 2} \int sin^{2-2} (x) dx -{1 \above{2pt} 2} sin^{2-1}(x) cos (x)$ simplifying;

$={1 \above{2pt} 2} \int sin^0 (x) dx -{1 \above{2pt} 2} sin(x) cos (x)$

$={1 \above{2pt} 2} \int 1 dx -{1 \above{2pt} 2} sin(x) cos (x)$

Now solving;

$\int 1 dx = x$ (Applying the constant rule)

$\implies {1 \above{2pt} 2} \int 1 dx -{1 \above{2pt} 2} sin(x) cos (x) = {x \above{2pt} 2} - {sin(x) cos (x) \above{2pt} 2}$

And therefore,

$\int sin^2(x)dx = {x \above{2pt} 2} - {sin(x) cos (x) \above{2pt} 2} + C$ Simplifying this;

we know that $sin (2x) = 2sin(x)cos(x)$

therefore, $sin(x)cos(x) = {1 \above{2pt} 2}sin(2x)$

$\therefore \int sin^2(x)dx = {x \above{2pt} 2} - {{1\above{2pt} 2} sin(2x) \above{2pt} 2} + C$

$={x \above{2pt} 2} - { sin(2x) \above{2pt} 4} + C$

$=-{sin (2x) - 2x \above{2pt} 4} + C$