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# Evaluate$\int {{{\sin }^4}x{{\cos }^3}xdx.}$  Answer Verified
Hint: Use substitution method i.e. substitute $\sin x = t$ for easy simplification.

Let, $I = \int {{{\sin }^4}x{{\cos }^3}xdx} = \int {{{\sin }^4}x{{\cos }^2}x\cos xdx}$
As we know${\cos ^2}x = \left( {1 - {{\sin }^2}x} \right)$, so substitute this value.
$\Rightarrow I = \int {{{\sin }^4}x{{\cos }^3}xdx} = \int {{{\sin }^4}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} .................\left( 1 \right)$
Now, let $\sin x = t$
Differentiate above equation w.r.t.$x$
As we know $\sin x$differentiation is $\cos x$
$\Rightarrow \cos xdx = dt$
So, substitute this value in equation (1).
$I = \int {{{\sin }^4}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} = \int {{t^4}\left( {1 - {t^2}} \right)dt} \\ \Rightarrow I = \int {\left( {{t^4} - {t^6}} \right)dt} \\$
Now, integrate it, as we know$\int {{t^n}dt = \left[ {\dfrac{{{t^{n + 1}}}}{{n + 1}}} \right]}$, so apply this property.
$\Rightarrow I = \left[ {\dfrac{{{t^5}}}{5} - \dfrac{{{t^7}}}{7}} \right] + c$, (where c is some arbitrary integration constant)
L.C.M of 5 and 7 is 35, and take ${t^5}$as common
$\Rightarrow I = \dfrac{{{t^5}}}{{35}}\left[ {7 - 5{t^2}} \right] + c$
Now re-substitute the value of $t = \sin x$
$\Rightarrow I = \dfrac{{\left( {{{\sin }^5}x} \right)}}{{35}}\left[ {7 - 5\left( {{{\sin }^2}x} \right)} \right] + c$
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 $\sin x = t$, 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.
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