Let the integral be:∫sin4(x)cos3(x)dxI have to integrate this function by changing the variable. I'm trying: u=sin⁡(x) and so du=cos⁡(x)dx. By rewriting the integral I get:∫u4cos2(x)duBut I'm stuck here because I'm not sure there should be any expression with x left in the integral.Also I know the final answer is :∫sin4xcos3xdx=−sin7x7+sin5x5

Question

Let the integral be:∫sin4(x)cos3(x)dxI have to integrate this function by changing the variable. I&#039;m trying: u=sin⁡(x) and so du=cos⁡(x)dx. By rewriting the integral I get:∫u4cos2(x)duBut I&#039;m stuck here because I&#039;m not sure there should be any expression with x left in the integral.Also I know the final answer is :∫sin4xcos3xdx=−sin7x7+sin5x5

Samantha Brown · Accepted Answer

Hint:Use the identitycos2(x)=1−sin2(x)and don&#039;t forget the constant of integration!

Toni Scott · Accepted Answer

As the exponent of cos is odd, you can set  u=sin⁡x,du=cos⁡xdx  If the function to integrate had been, say sin3xcos4xdx, we would have set  u=cos⁡x,du=−sin⁡xdx

Vasquez · Accepted Answer

∫sin4⁡xcos3⁡xdx=−∫−15cos2⁡x(−5cos⁡(x)sin4⁡x)dx=−∫−15cos2⁡x(sin5⁡x)′dxNow perform an integration by parts and use the same technique for the new integral.

Let the integral be: \int \sin^4(x)\cos^3(x)dx NS

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