Question

Evaluate: \[\int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x) + 3\cos (\dfrac{1}{2}\cos x))\sin xdx} \]

Answer 1

Hint:
We will apply integration to both the terms. We will write it separately for both the terms. Then, we will assume \[\cos \,x\] as t. Then, we will write the limit in terms of t. And then we will integrate. Then, we will get the answer.

Complete step by step solution:
Given,
 \[\int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x) + 3\cos (\dfrac{1}{2}\cos x))\sin xdx} \]
We can also write it as
 \[\int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x) + 3\cos (\dfrac{1}{2}\cos x))\sin xdx} = \int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x))\sin xdx + \int\limits_0^\pi {3\cos (\dfrac{1}{2}\cos x)\sin xdx} } \] …. (1)
Here we will use the following formula
 \[\int\limits_0^{2a} {f(x)dx = \left\{ {
  2\int\limits_0^a {f(x)dx} ,f(2a - x) = 2a \\
  0, f(2a - x) = - f(x) } \right\}} \]
Using this formula, the first term in equation (1) will become zero.
 \[ \Rightarrow \int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x))\sin xdx = 0} \]
Therefore, we have
 \[ \Rightarrow \int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x) + 3\cos (\dfrac{1}{2}\cos x))\sin xdx = } \int\limits_0^\pi {{e^{\cos x}}3\cos (\dfrac{1}{2}\cos x))\sin xdx} \]
This can also be written as
 \[ = 2\int\limits_0^{\dfrac{\pi }{2}} {{e^{\cos x}}3\cos (\dfrac{1}{2}\cos x))\sin xdx} \] …. (2)
Let \[\cos x = t\]
 \[ \Rightarrow - \sin xdx = dt\]
At \[x = 0,t = 1\]
And at \[x = \dfrac{\pi }{2},t = 0\]
Putting these values in equation (2), we have
 \[ = 2\int\limits_0^1 {{e^t}3\cos (\dfrac{1}{2}t))dt} \]
Taking 3 outside the integration
 \[ = 6\int\limits_0^1 {{e^t}\cos (\dfrac{1}{2}t))dt} \]
Here we will use the formula
 \[\int {{e^{ax}}\cos bxdx = \dfrac{{{e^{ax}}}}{{{a^2} + {b^2}}}(b\sin bx + a\cos bx)} \]
On integrating, we
 \[ = 6\left[ {\dfrac{{{e^t}}}{{1 + \dfrac{1}{4}}}\left( {\dfrac{1}{2}\sin \dfrac{x}{2} + \cos \dfrac{x}{2}} \right)} \right]_0^1\]
Putting the values of limit
 \[ = \dfrac{{24}}{5}\left[ {e\left( {\dfrac{1}{2}\sin \dfrac{1}{2} + \cos \dfrac{1}{2}} \right) - {e^0}\left( {\cos 0} \right)} \right]\]
Since the value of \[\cos 0\] is 1 and the value of \[{e^0}\] is 1.
Therefore, we have
 \[ = \dfrac{{24}}{5}\left[ {\dfrac{e}{2}\sin \dfrac{1}{2} + e\cos \dfrac{1}{2} - 1} \right]\]

Hence, the value of \[\int\limits_0^\pi {{e^{\left| {\cos x} \right|}}(2\sin (\dfrac{1}{2}\cos x) + 3\cos (\dfrac{1}{2}\cos x))\sin xdx} = \dfrac{{24}}{5}\left[ {\dfrac{e}{2}\sin \dfrac{1}{2} + e\cos \dfrac{1}{2} - 1} \right]\].

Note:
Definite integrals are the integrals which have upper and lower limits. Here, we have used the properties of definite integrals.
We know that,
 \[
    \\
  \int\limits_0^{2a} {f(x)dx = } \int\limits_0^a {f(x)dx + } \int\limits_0^a {f(2a - x)dx} \\
 \]
If \[f(2a - x) = f(x)\]. Then, \[\int\limits_0^{2a} {f(x)dx = } \int\limits_0^a {f(x)dx + } \int\limits_0^a {f(x)dx} = 2\int\limits_0^a {f(x)dx} \]
If \[f(2a - x) = - f(x)\]. Then, \[\int\limits_0^{2a} {f(x)dx = } \int\limits_0^a {f(x)dx - } \int\limits_0^a {f(x)dx} = 0\]
 \[ \Rightarrow \int\limits_0^{2a} {f(x)dx = \left\{ {
  2\int\limits_0^a {f(x)dx} ,f(2a - x) = 2a \\
  0,f(2a - x) = - f(x) } \right\}} \]