Question

integration of (sinx)^2
Write down the process pls

Answer 1

Answer:

we know that

cos2x=1-2sin^2x

so,

sin^2x=(1-cos2x)/2

integration sin^2x=integration ((1-cos2x)/2)

=x/2-sin2x/4

Answer 2

Answer:

Step-by-step explanation:

First let us simplify (sin x)^2

(sin x)^2 = $sin^{2}x$

We know that

cos2x = 1 - 2$sin ^{2} x$

-2$sin ^{2} x$ = cos 2x -1

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

2$sin ^{2} x$ = -cos 2x + 1

2$sin ^{2} x$ = 1 - cos 2x

$sin ^{2} x$  = $\frac{1 - cos 2x }{2}$

Now

∫$sin ^{2} x$ dx = ∫$\frac{1 - cos 2x }{2}$ dx

                = $\frac{1}{2}$ ∫ 1 - cos 2x dx                   {$\frac{1}{2}$ is constant so we took it out}

                = $\frac{1}{2}$ [∫ 1 dx - ∫ cos 2x dx] + C     {C is constant}      

                = $\frac{1}{2}$ [x - $\frac{sin 2x}{2}$] + C                 {Integration of 1 is x and integration of                                                                                

                                                                         cos 2x is  $\frac{sin 2x}{2}$  }

∫$sin ^{2} x$ dx  =  $\frac{x}{2} - \frac{sin2x}{4}$ + C

Therefore  integration  of  (sin x)^2  is  $\frac{x}{2} - \frac{sin2x}{4}$ + C

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