How do we Calculate the following type of Questions :-

1)When the Trignometric Function has power like - Integration of 

2)Find area bound by curve like-y=|x-1| and y=3-|X|

k no beatin abt d bush

1) by practice of course

2) draw d grafs to get an idea nd proceed using simple geometeical methods or integration if required....if itz mod u wont need much of integration....can b solved by simpl geometry

(sinx)^4=(1/2-cos2x/2)^2

              =1/4-cos2x/2+(cos2x)^2/4

              =1/4-cos2x/2+1/8+cos4x/8

              =3x/8-sin2x/4+sin4x/32+c

Please tell whether my solution is correct or not?

certain other types

pattern of indefinite integration for sin 2nx/ sin x



sin 2x/ sin x = 2 [ sin x] + C



sin 4x/ sin x = 2[ sin x + (sin 3x) / 3 ] + c



sin 6x/ sin x = 2[ sin x + (sin 3x) / 3 + (sin 5x) / 5] + c



in this way we can integrate even sin10x/sinx, sin 12x/sinx, etc. the value of



sin 10x/ sin x = 2[ sin x + (sin 3x) / 3 + (sin 5x) / 5 + (sin 7x) / 7 + (sin9x) / 9] + c

for odd

we know,



_{[ ]}^{[ ]} sinx dx / sin x  = x + C





_{[ ]}^{[ ]} sin 3x dx / sin x  = x + sin 2x + C





_{[ ]}^{[ ]} sin 5x dx / sin x  = x + sin 2x + sin 4x / 4 + C





therefore,



_{[ ]}^{[ ]} sin 7x dx / sin x =  x + sin 2x + sin 4x / 2  +  sin 6x / 3 + C





in this method, we can find integration of sin 9x / sin x , sin 11x/ sin x etc.

Proof :

let us take _{[ ]}^{[ ]} sin7x/sinx

 we know    sin7x = sin ( 6x +x )

sin 7x = sin6x cos x + cos 6x sinx

sin7x / sinx = ( sin6x cos x + cos 6x sinx ) / sinx

                  = ( sin6x cosx / sinx)   + cos 6x             

now, ( sin6x cosx / sin x)  = 2sin6x cosx / 2sinx

                                       = ( sin  7x + sin 5x ) / 2sinx

                                       = ( sin  7x / 2sinx ) + ( sin 5x / 2sinx )



so, sin7x / sinx = ( sin  7x / 2sinx ) + ( sin 5x / 2sinx ) + cos 6x

so, sin 7x / 2sinx = ( sin 5x / 2sinx ) + cos 6x

so, sin7x / sinx = ( sin 5x / sinx ) + 2 cos 6x

again,



sin 5x / sinx = ( sin x cos 4x + cos x sin 4x ) / sin x

                   = ( sin x cos 4x + 4 sin x cos x cos 2x cos x ) / sin x

                  = cos 4x + 4 cos x  cos 2x cos x

                 = cos 4x + 2 cos x (  2 cos 2x cos x )

                 = cos 4x + 2 cos x ( cos 3x + cos x )

                 = cos 4x + 2 cos x cos 3x + 2 cos ² x

               = cos 4x + cos 4x + cos 2x + cos 2x + 1

              = 2 cos 4x + 2 cos 2x + 1

so,



sin7x / sinx = 2 cos 4x + 2 cos 2x + 1 + 2 cos 6x

now integrating both sides,

answer is

  _{[ ]}^{[ ]} sin7x/sinx  =  sin4x / 2 + sin6x / 3 + sin2x  + x +  C

can you tell me whether it is (sinx)^4 or not?