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Evaluatesin4xcos3xdx.

Answer
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Hint: Use substitution method i.e. substitute sinx=t for easy simplification.

Let, I=sin4xcos3xdx=sin4xcos2xcosxdx
As we knowcos2x=(1sin2x), so substitute this value.
I=sin4xcos3xdx=sin4x(1sin2x)cosxdx.................(1)
Now, let sinx=t
Differentiate above equation w.r.t.x
As we know sinxdifferentiation is cosx
cosxdx=dt
So, substitute this value in equation (1).
I=sin4x(1sin2x)cosxdx=t4(1t2)dtI=(t4t6)dt
Now, integrate it, as we knowtndt=[tn+1n+1], so apply this property.
I=[t55t77]+c, (where c is some arbitrary integration constant)
L.C.M of 5 and 7 is 35, and take t5as common
I=t535[75t2]+c
Now re-substitute the value of t=sinx
I=(sin5x)35[75(sin2x)]+c
So, this is the required value of the integration.

Note: In such types of questions always choose substitution which makes integration simple, in above integration we choose sinx=t, so it makes integration simple, then we easily integrate using some basic property of integration which is stated above, then simplify we will get the required answer.