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`        int of (a)  2^x*e^x dx, int of (b)  4*cosx/2*cosx*sin21x/2 dx, int of (c)  cos2x-cos2y/cosx-cosy dx, int of (d)  [sinx*sin(y-x)+sin^2(y/2-x)] dx, int of (e)  sin2x+sin5x-sin3x/cosx+1-2sin^2x dx, int of (f)  cos8x-cos7x/1+2cos5x dx`
8 years ago

Ramesh V
70 Points
```										using the  relationsSum / Difference of Trigonometric Functions Formulas.   1.  sin A + sin B = 2 sin [ (A + B) / 2 ] cos [ (A - B) / 2 ]   2. sin A - sin B = 2 cos [ (A + B) / 2 ] sin [ (A - B) / 2 ]   3. cos A + cos B = 2 cos [ (A + B) / 2 ] cos [ (A - B) / 2 ]    4.  cos A - cos B = - 2 sin [ (A + B) / 2 ] sin [ (A - B) / 2 ]a) let A= 2x*ex.dxintegrating by parts givesA = 2x*ex - ln 2 Aso A = 2x*ex / ln(2e) + C   where C is constantb) Question not clearc)  cos2x-cos2y/cosx-cosy dx   = (2 cos2x-1-2cos2y+1)/(cosx-cosy) dx   = 2 (cosx+cosy) dx    = 2 (sinx+x.cosy)+ C   where C is constantd) [sinx*sin(y-x)+sin^2(y/2-x)] dx  = using relation 4 stated above , we can write as   = 1/2*( cos(2x-y) - cos y ) + 1/2* (1-cos(2x-y) ) dx   here cos 2x = 1 - 2sin2x  so we have finally  = 1/2 * (1-cosy).dx  = 1/2 * (1-cosy) + C where C is constante) (sin2x+sin5x-sin3x)/(cosx+1-2sin2x) dx  here here cos 2x = 1 - 2sin2x   = (sin2x+sin5x-sin3x)/(cosx+cos 4x) dx   using relations 1 and 3 we have   =  (-2cox 5x/2. sin x/2+2 sin5x/2 cos5x/2) / (2.cos 5x/2.cos3x/2) dx  =  ( sin x/2 - sin x/2) / (cos3x/2) dx  =  (2cox 3x/2. sin x) / (cos3x/2) dx  =  (2sin x) dx  =  - 2 cos x  + C where C is constantf)  (cos8x-cos7x)/(1+2cos5x) dx   Adding +cos 2x and - cos 2x to numerator gives         ={(cos8x+cos 2x)-(cos7x+cos 2x)}/(1+2cos5x) dx       = {(2cos5xcos 3x - (cos7x+cos 2x)}/(1+2cos5x) dx       again adding +cos 3x and - cos 3x to numerator gives       = {(2cos5xcos 3x + cos 3x - (cos7x+cos 2x+cos 3x)}/(1+2cos5x) dx      ={(2cos5xcos 3x + cos 3x - (2 cos 5x.cos 2x+cos 2x)}/(1+2cos5x) dx      ={(2cos5x+1)cos 3x - (2 cos 5x+1).cos 2x}/(1+2cos5x) dx     =(cos 3x  - cos 2x) dxAnswer is 1/3*sin 3x - 1/2*sin 2x +C     where C is constant
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8 years ago
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