Integral of Sin X Squared

 The integral calculus deals with the inverse problem of the derivative of the given function. If F(x) = sin x is an function and the derivative of function with respect to x is f(x) = cos x , then we say that the integral of f(x) with respect  to x is F(x). In trigonometric term like sin, cos, tan also can be performed in integral calculus.  

                                  i.e.,  `int` cos x dx = sin x + c,         where c is constant.

 

Integral Formulas on Integral of Sin X Squared:

 

1. ` int` x n dx = `(x^n+1) / (n+1)`

2.` int ` cos x .dx = sin x. + c

3. `int` sin x.dx = - cos x + c

4. `int` sec2x. dx = tan x + c

5.` int` cosec x.cot x. dx = -cosec x + c

6. `int (dx) / (x^2 + a^2)` =` (1/a) tan^-1(x / a) + c`

7. `int (dx) / (a^2 - x^2) ` = `(1/(2a)) log [(a + x) / (a - x)] + c`

8. `int (dx) / (x^2 - a^2)` = `(1/(2a)) log [(x - a) / (x + a)] + c`

9. `int (dx) / sqrt(a^2 - x^2)` = `sin^-1(x / a) + c`

10`int (dx) / sqrt(x^2 - a^2) ` = `log [(x + sqrt(x^2 - a^2)] + c`

11.  `int` e x dx = e x + c

12. `int` u dv = uv -`int` v du

 

Problems on Integral of Sin X Squared

 

Let us see some examples on Integral of sin x squard.

Integral of sin x squared problem 1:

    Find the integration of  `int` sin2x dx

    Solution:

         `int` sin2x dx  = `int` sin2x dx                                                 

        we know the relation between the trigonometric term   sin2x =`(1/2)` (1 - cos 2x)

                                = `int` `(1 - cos 2x)/2` dx

                                = `int` `(dx)/2 ` - `int ` `(cos 2x)/2 ` dx

                                = `1/2` ` int` dx - `1/2 ` `int` cos 2x dx

                                = `1/2` x - `1/2` `. (1/2) ` sin 2x + c

                                =  ` x/2` - `(1/4)` sin 2x + c

     Answer:     ` x/2` - `(1/4)` sin 2x + c

 Integral of sin x squared problem 2:

       Find the integration of  sin3 x . cosx

   Solution:.

                         ` int ` sin3 x . cosxdx      

                           ` int ` sin3 x . cosx dx  =` int `sin2 x . cosx (sin x) dx             we know  1-cosx = sin2 x

                                                                =`int` (1-cosx) cosx (sinx) dx      

Put t = cos x   so thet dt = -sin x dx

Therefore,     `int` (1-cosx) cosx (sinx) dx = ` - ` `int` (1-t2) t2 dt

                                                                   = ` -` `int ` (t2 - t4 )dt

                                                                   =  `- ( t^3/3 - t^5/5)+c`

                                                                   = `-1/3 cos^3x ` + `1/5 cos^5x` + c

Answer:  `-1/3 cos^3x ` + `1/5 cos^5x` + c.