Question

How do you find the integral of $\int {{{\sin }^n}(x){{\cos }^m}(x)}$ if $m$ and $n$ are integers?

Answer 1

Hint : Here, since m and n are integers four different cases arise based on whether and are even or odd. We will use substitution and trigonometric identities to either write sine in terms of cosine or vice-versa, to get an integrable reduced form of the integral. We must know some of the basic trigonometric identities to solve the given integral.

Complete step-by-step answer :
We can see that the given integral has the powers of sine and cosine. So, for such problems we will use the following guidelines to reduce the integral and then integrate it using the known methods. These guidelines have to be followed when evaluating $\int {{{\sin }^n}(x){{\cos }^m}(x)}$, where $m \geqslant 0{\text{ and }}n \geqslant 0$ are integers. We encounter four different cases here.

Case (i): If the power of cosine function $m$ is odd, that is $m = 2k + 1$ , we take one cosine out of the ${\cos ^m}(x)$ and use the identity ${\cos ^2}x = 1 - {\sin ^2}x$ to express the remaining cosine in terms of sine.
$\Rightarrow \int {{{\sin }^n}(x){{\cos }^m}(x)} dx = \int {{{\sin }^n}(x){{\left( {{{\cos }^2}(x)} \right)}^k}\cos (x)} dx$
$\Rightarrow \int {{{\sin }^n}(x){{\cos }^m}(x)} dx = \int {{{\sin }^n}(x){{\left( {1 - {{\sin }^2}(x)} \right)}^k}\cos (x)} dx$ - - - - - - - - - - - - - - - - (1)
Then use the substitution $u = \sin (x)$.
$\Rightarrow \dfrac{{du}}{{dx}} = \cos (x)$
$\Rightarrow du = \cos (x)dx$
So, (1) becomes $\int {{{\sin }^n}(x){{\cos }^m}(x)} dx = \int {{u^n}{{\left( {1 - {u^2}} \right)}^k}du}$, which can be easily evaluated.

Case (ii): Similarly, if the power of sine function $n$ is odd, that is $n = 2k + 1$ , we take one sign out of the ${\sin ^m}(x)$ and use the identity ${\sin ^2}x = 1 - {\cos ^2}x$ to express the remaining sine in terms of cosine.
$\Rightarrow \int {{{\sin }^n}(x){{\cos }^m}(x)} dx = \int {{{\left( {{{\sin }^2}(x)} \right)}^k}\sin (x){{\cos }^m}(x)} dx$
$\Rightarrow \int {{{\sin }^n}(x){{\cos }^m}(x)} dx = \int {{{\left( {1 - {{\cos }^2}(x)} \right)}^k}\sin (x)} {\cos ^m}(x)dx$ - - - - - - - - - - - - - - - - (1)
Then use the substitution $u = \cos (x)$.
$\Rightarrow \dfrac{{du}}{{dx}} = - \sin (x)$
$\Rightarrow - du = \sin (x)dx$
So, (1) becomes $\int {{{\sin }^n}(x){{\cos }^m}(x)} dx = - \int {{u^m}{{\left( {1 - {u^2}} \right)}^k}du}$, which can be easily evaluated.

Case(iii): When both the powers $n{\text{ and }}m$ are odd, any one of the above methods can be used.

Case(iv): When both the powers $n{\text{ and }}m$ are even, the method of solving each of the integrals differ. We will generally use the half-angle identities ${\sin ^2}x = \dfrac{1}{2}\left( {1 - \cos \left( {2x} \right)} \right)$ and ${\cos ^2}x = \dfrac{1}{2}\left( {1 + \cos \left( {2x} \right)} \right)$ . The identity $\sin x\cos x = \dfrac{1}{2}\sin \left( {2x} \right)$ is also used sometimes. In general, when the powers are even, we will use a series of these formulas to reduce the integral to form which we know to integrate.
So, by using the above techniques we can easily reduce the integral to an easily integrable form and find its answer. Hence, the integral $\int {{{\sin }^n}(x){{\cos }^m}(x)}$ can be evaluated.

Note : Note that, we will face situations where other trigonometric identities also have to be used in order to reduce the integral in the case (iv). In all other cases the reduced form is easier to get. We must know the integration by substitution method to solve the given problem.