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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
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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