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

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

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,

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\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, $$\int\limits_0^{2a} {f(x)dx = } \int\limits_0^a {f(x)dx - } \int\limits_0^a {f(x)dx} = 0$$ $$ \Rightarrow \int\limits_0^{2a} {f(x)dx = \left\\{ { 2\int\limits_0^a {f(x)dx} ,f(2a - x) = 2a \\\ 0,f(2a - x) = - f(x) } \right\\}} $$