 QUESTION

# Integrate the following expression : $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x\cos xdx}$.

Hint: We will proceed with the assumption that $u=\sin x$, then we get $du=\cos xdx$ or $dx=\dfrac{du}{\cos x}$, then we will put the value of $dx$ in the main integration to solve it further.

We have to calculate the expression $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x\cos xdx}$.
Here we have a fixed limit from 0 to $\dfrac{\pi }{2}$ and we know that both $\sin x$ and $\cos x$ are positive in the first quadrant.
Let us assume that $u=\sin x$ then differentiating $u=\sin x$ with respect to x,
We get- $\dfrac{du}{dx}=\cos x$ $du=\cos xdx$ or, $dx=\dfrac{du}{\cos x}$,
On putting the value of $\sin x$ and $dx$ in the given equation,
We get, $\int\limits_{0}^{\dfrac{\pi }{2}}{u\times \cos x\times \dfrac{du}{\cos x}}$,
Now the value of the limit also changes as $\sin 0=0$ and $\sin \dfrac{\pi }{2}=1$.
So, our lower limit will be 0 and upper limit will be 1.
We get the expression,
$\int\limits_{0}^{\dfrac{\pi }{2}}{u\times \cos x\times \dfrac{du}{\cos x}}$ modified as $\int\limits_{0}^{1}{u\times du}$.
Now, we know that $\int{ndn=\dfrac{{{n}^{2}}}{2}}$, so we get
$\int\limits_{0}^{1}{u\times du}=\left[ \dfrac{{{u}^{2}}}{2} \right]_{0}^{1}$.
Evaluating the limits, we get
= $\left[ \dfrac{{{1}^{2}}}{2}-\dfrac{{{0}^{2}}}{2} \right]$ = $\dfrac{1}{2}$ .
Therefore the integration of expression $\int\limits_{0}^{\dfrac{\pi }{2}}{\sin x\cos xdx}=\dfrac{1}{2}$.

Note: We can also solve this question by assuming $u=\cos x$, then we get $\dfrac{du}{dx}=-\sin x$ or $dx=\dfrac{du}{-\sin x}$ . On putting the value of these two equations in the expression given and following the same steps , changing the limits and solving the definite integral so formed, as in the solution above will result in the same value. This is possible because both the sine and cosine trigonometry functions are complementary to each other. Other trigonometry functions that are complementary to each other are - tangent and cotangent, secant and cosecant.