Engineering Mathematics Test 6

\(\underset{-\pi }{\overset{\pi }{\mathop \int }}\,cos\left( t \right)\cos \left( 3t \right)dt\)

We know that, 2 cos A cos B = cos (A + B) + cos (A – B)

\(=\frac{1}{2}\underset{-\pi }{\overset{\pi }{\mathop \int }}\,\left[ \cos \left( 4t \right)+\cos \left( -2t \right) \right]dt\)

\(=\frac{1}{2}\left[ \frac{\sin 4t}{4}+\frac{\sin 2t}{2} \right]_{-\pi }^{\pi }=0\)

\(f = \mathop \smallint \limits_0^{\sqrt {\frac{\pi }{2}} } xsin({x^2})\;dx\)

\(put\;t = \;{x^2}\)

\(dt = 2xdx\)

\(\therefore \frac{{dt}}{2} = xdx\)

\(when\;x = 0\;then\;t = 0\)

\(when\;x = \sqrt {\frac{\pi }{2}} \;then\;t = \frac{\pi }{2}\)

\(f = \mathop \smallint \limits_0^{\frac{\pi }{2}} \left( {\frac{{{\rm{sin}}\left( t \right)dt}}{2}} \right)\)

\(f = \; - \frac{1}{2}\left[ {\cos \left( {\frac{\pi }{2}} \right) - \cos (0)\;} \right]\)

\(f = \; - \frac{1}{2}\left[ {0\; - 1} \right]\)

\(f = \;\frac{1}{2}\)

Let \(I = \mathop \smallint \limits_0^3 \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^{2/3}}}}dx\)

The given integral is undefined at x = 1. So, we can split the above integral as follows.

\(= \mathop \smallint \limits_0^1 \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^{2/3}}}}dx + \mathop \smallint \limits_1^2 \frac{{2x}}{{{{\left( {1 - {x^2}} \right)}^{2/3}}}}dx\)

\(= \left[ { - 3{{\left( {1 - {x^2}} \right)}^{1/3}}} \right]_0^1 + \left[ { - 3{{\left( {1 - {x^2}} \right)}^{1/3}}} \right]_1^3\)

\(\begin{array}{l} \mathop \smallint \limits_a^b f\left( x \right)dx = \mathop \smallint \limits_a^b f\left( {a + b - x} \right)dx\\ I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx\;\_\_\_\left( 1 \right) \end{array}\)

\(I = \mathop \smallint \limits_a^{\frac{\pi }{2}} \frac{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} }}{{\sqrt {\sin \left( {\frac{\pi }{2} - x} \right)} + \sqrt {\cos \left( {\frac{\pi }{2} - x} \right)} }}d\)

\(I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{\sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}dx\;\_\_\left( 2 \right)\)

Adding both:

\(\begin{array}{l} 2I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\cos x} + \sqrt {\sin x} }}dx = \mathop \smallint \limits_0^{\frac{\pi }{2}} 1\;dx = \frac{\pi }{2}\\ I = \frac{\pi }{4} \end{array}\)  

\(I = \mathop \smallint \limits_0^\infty \frac{{dx}}{{{{\left( {{x^2} + 1} \right)}^2}}}\)

Put x = tan θ

⇒ dx = sec² θ dθ

Limits will be: 0 to π/2

\(I = \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{{{\sec }^2}\theta }}{{{{\left( {ta{n^2}\theta + 1} \right)}^2}}}d\theta\)

\(= \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{{{\sec }^2}\theta }}{{{{\left( {{{\sec }^2}\theta } \right)}^2}}}d\theta\)

\(= \mathop \smallint \limits_0^{\frac{\pi }{2}} {\cos ^2}\theta d\theta\)

\(= \mathop \smallint \limits_0^{\frac{\pi }{2}} \frac{{1 + \cos 2\theta }}{2}d\theta\)

Concept:

\(\;\mathop \smallint \nolimits_0^{\frac{\pi }{2}} {\sin ^n}xdx = \mathop \smallint \nolimits_0^{\frac{\pi }{2}} {\cos ^n}x\;dx \)

If n is even, \(\frac{{\left( {n - 1} \right)\left( {n - 3} \right)\left( {n - 5} \right) \ldots \ldots }}{{n\left( {n - 2} \right)\left( {n - 4} \right)\left( {n - 6} \right) \ldots ..}} \times \frac{\pi }{2}\;\)

If n is odd, \(\frac{{\left( {n - 1} \right)\left( {n - 3} \right)\left( {n - 5} \right)}}{{n\left( {n - 2} \right)\left( {n - 4} \right)\left( {n - 6} \right)}} \times 1\)                                            
Here n = 7, means n is odd,

Therefore, \(\mathop \smallint \nolimits_0^{\frac{\pi }{2}} {\sin ^9}xdx = \;\frac{{\left( {9 - 1} \right)\left( {9 - 3} \right)\left( {9 - 5})({9-7} \right)}}{{9\left( {9 - 2} \right)\left( {9 - 4} \right)\left( {9 - 6} \right)(9-8)}} \times 1 = 0.2031\)

Explanation:

\({\rm{f}}\left( {\rm{x}} \right) = {\rm{Acos}}\left( {\frac{{2{\rm{\pi x}}}}{5}} \right) + {\rm{C}}\)

\({\rm{f'}}\left( {\rm{x}} \right) = - {\rm{Asin}}\left( {\frac{{2{\rm{\pi x}}}}{5}} \right) \times \frac{{2{\rm{\pi }}}}{5}\)

\({\rm{f'}}\left( {\frac{{15}}{4}} \right) = - {\rm{Asin}}\left( {\frac{{2{\rm{\pi }}\left( {\frac{{15}}{4}} \right){\rm{\;}}}}{5}} \right) \times \frac{{2{\rm{\pi }}}}{5}\)

\({\rm{f'}}\left( {\frac{{15}}{4}} \right) = - {\rm{A}}\left( { - 1} \right)\left( {\frac{{2{\rm{\pi }}}}{5}} \right) = \frac{1}{2}{\rm{\;}}\)

\(\therefore {\rm{A}} = \frac{5}{{4{\rm{\pi }}}}\)

\(\mathop \smallint \nolimits_0^5 {\rm{f}}\left( {\rm{x}} \right){\rm{dx}} = \frac{{A\pi }}{2}\)

\(f\left( x \right) = A~\cos\left( {\frac{{2\pi x}}{5}} \right) + C\)

\(\mathop \smallint \nolimits_0^5 \left( {\;\frac{5}{{4{\rm{\pi }}}}{\rm{cos}}\left( {\frac{{2{\rm{\pi x}}}}{5}} \right) + {\rm{C\;}}} \right){\rm{dx\;}} = \frac{{\left( {\frac{5}{{4\pi }}} \right)\pi }}{2}\)

\(\left[ {\frac{5}{{4\pi }}\left\{ {\sin \left( {\frac{{2\pi x}}{5}} \right) \times \frac{5}{{2\pi }}} \right\} + Cx} \right]_0^5 = \frac{5}{8}\)

\(5C = \frac{5}{8}\)

\(C = \frac{1}{8}\)

Property:

\(\mathop{\int }_{\text{a}}^{\text{b}}\text{f}\left( \text{x} \right)\text{ }\!\!~\!\!\text{ dx }\!\!~\!\!\text{ }=\mathop{\int }_{\text{a}}^{\text{b}}\text{f}\left( \text{a}+\text{b}-\text{x} \right)\text{ }\!\!~\!\!\text{ dx}\)

Calculation:​​​

\(\text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{x }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\left( \text{ }\!\!\pi\!\!\text{ }-\text{x} \right)\text{ }\!\!~\!\!\text{ }{{\left( \cos \left( \text{ }\!\!\pi\!\!\text{ }-\text{x} \right) \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}\)

\(\text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}-\text{ }\!\!~\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{x }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}\)

\(\text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}-\text{ }\!\!~\!\!\text{ I}\)

\(2\times \text{I}=\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\text{ }\!\!\pi\!\!\text{ }\!\!~\!\!\text{ }{{\left( \cos \text{x} \right)}^{2}}\text{ }\!\!~\!\!\text{ dx}=\text{ }\!\!\pi\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\frac{1+\cos 2\text{x}}{2}\text{ }\!\!~\!\!\text{ dx}\)

\(2\times \text{I}=\text{ }\!\!\pi\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\frac{1}{2}\text{ }\!\!~\!\!\text{ dx}+\text{ }\!\!\pi\!\!\text{ }\mathop{\int }_{0}^{\text{ }\!\!\pi\!\!\text{ }}\frac{\cos 2\text{x}}{2}\text{ }\!\!~\!\!\text{ dx}\)

\(2\times \text{I}=\text{ }\!\!~\!\!\text{ }\frac{\text{ }\!\!\pi\!\!\text{ }}{2}\left( \text{ }\!\!\pi\!\!\text{ }-0 \right)+\frac{1}{2}\left( \sin 2\text{ }\!\!\pi\!\!\text{ }-\sin 0 \right)=\frac{{{\text{ }\!\!\pi\!\!\text{ }}^{2}}}{2}\)

\(\therefore \mathbf{I}=\mathop{\int }_{0}^{\mathbf{\pi }}\mathbf{x}~{{\left( \cos \mathbf{x} \right)}^{2}}~\mathbf{dx}=\frac{{{\mathbf{\pi }}^{2}}}{4}\)

\(\mathop \smallint \limits_0^1 \frac{{{e^t}}}{{{{\left( {1 + t} \right)}^2}}}dt\)

By integrating by parts

\( = {e^t}\mathop \smallint \limits_0^1 \frac{{dt}}{{{{\left( {1 + t} \right)}^2}}} - \mathop \smallint \limits_0^1 {\left( {{e^t}.\smallint \frac{{dt}}{{{{\left( {1 + t} \right)}^2}}}} \right)^2}dt\)

\( = {e^t}\mathop \smallint \limits_0^1 \frac{{dt}}{{{{\left( {1 + t} \right)}^2}}} - \mathop \smallint \limits_0^1 {e^t}.\left( {\frac{{ - 1}}{{1 + t}}} \right)dt\)

Given that,

\(\mathop \smallint \limits_0^1 {e^t}\left( {\frac{{ 1}}{{1 + t}}} \right)dt = a\)

\( = \left[ {{e^t}.\frac{{ - 1}}{{\left( {1 + t} \right)}}} \right]_0^1 + a\)

\( = 1 - \frac{e}{2} + a\)

\(let\;I = \mathop \smallint \limits_0^{\frac{\pi }{2}} {\sin ^7}x{\cos ^5}xdx\)

\(It\;is\;of\;the\;form:\mathop \smallint \limits_0^{\frac{\pi }{2}} {\sin ^m}x{\cos ^{\rm{n}}}xdx\)

\(I = \frac{{\left( {m - 1} \right)\left( {m - 3} \right)\left( {m\; - \;5} \right) \ldots 1\; \times \left( {n\; - 1} \right)\left( {n - 3} \right)\; \ldots 1}}{{\left( {m + n} \right)\left( {m + n - 2} \right)\left( {m + n - 4} \right) \ldots 1}}\)

\(I = \frac{{6 \times 4 \times 2 \times 4 \times 2}}{{12 \times 10 \times 8 \times 6 \times 4 \times 2}}\)

\(I = \frac{1}{{120}}\)