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Evaluate : (i) intsin3xsin2xdx (ii) intcos3xsin2xdx (iii) intcos4xcosxdx (iv) intsin^(3)xcos^(3)xdx |
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Answer» Solution :(i) Using 2 sin a sin b = cos (a-b)-cos(a+b), we have `intsin3xsin2xdx=(1)/(2)int2sin3xsin2xdx` `=(1)/(2)int(cosx-cos5x)DX` `=(1)/(2)intcosxdx-(1)/(2)cos5xdx` `=(1)/(2)sinx-(sin5x)/(10)+C`. (ii) Using 2 cos a sin b = sin (a+b)-sin(a-b), we get `intcos3xsin2xdx=(1)/(2)int2cos3xsin2xdx` `=(1)/(2)int(sin5x-sinx)dx` `=(1)/(2)intsin5xdx(1)/(2)intsinxdx` `=(-cos5x)/(10)+(cosx)/(2)+C`. (iii) Using 2 cos a cos b = cos (a+b)+cos(a-b), we get `intcos4xcosxdx=(1)/(2)int2cos4xcosxdx` `=(1)/(2)int(cos5x+cos3x)dx` `=(1)/(2)intcos5xdx+(1)/(2)intcos3xdx` `=(sin5x)/(10)+(sin3x)/(6)+C`. (iv) `intsin^(3)xcos^(3)xdx=intsin^(3)xcos^(2)xcosxdx` `=intsin^(3)x(1-sin^(2)x)cosxdx` `=intt^(3)(1-t^(2))DT," where"sinx=t` `=intt^(3)dt-intt^(5)dt=(t^(4))/(4)-(t^(6))/(6)+C` `=(1)/(4)sin^(4)-(1)/(6)sin^(6)x+C`. |
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