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This is one of our many quizzes on Introduction to integrals. Click on this to access more such competitive quizzes on Calculus.

Introduction to integrals Quiz 2

mathematics

 Integrate the following term: $I = \int {\dfrac{{{x^4}}}{{\sqrt {1 - {x^{10}}} }}dx} $.

 Replace ${x^5} = t$ and use substitution method also use inverse trigonometric integration $\left[ {\int {\dfrac{{dt}}{{\sqrt {1 - {t^2}} }} = {{\sin }^{ - 1}} + C} } \right]$.

$  I = \int {\dfrac{{{x^4}}}{{\sqrt {1 - {x^{10}}} }}dx}  \\   \Rightarrow \int {\dfrac{{{x^4}}}{{\sqrt {1 - {{\left( {{x^5}} \right)}^2}} }}dx}  \\  \left\{ {{x^5} = t \Rightarrow 5{x^4}dx = dt \Rightarrow {x^4}dx = \dfrac{{dt}}{5}} \right\} \\   \Rightarrow \int {\dfrac{1}{{5\sqrt {1 - {t^2}} }}dt}  \\   \Rightarrow \dfrac{1}{5}{\sin ^{ - 1}}t + C \\   \Rightarrow \dfrac{1}{5}{\sin ^{ - 1}}\left( {{x^5}} \right) + C \\  $

 Find the integration of: $I = \int {x\sqrt {6x + 5} } dx$.

 Replace $\sqrt {6x + 5}  = t$ and then solve using substitution method.

$  \int {x\sqrt {6x + 5} } dx \\   \Rightarrow 6x + 5 = t \\   \Rightarrow x = \dfrac{{t - 5}}{6} \\   \Rightarrow dx = dt \\   \Rightarrow \int {\left( {\dfrac{{t - 5}}{6}} \right)\sqrt t } dt \\   \Rightarrow \dfrac{1}{6}\int {\left( {{t^{\dfrac{3}{2}}} - 5{t^{\dfrac{1}{2}}}} \right)} dt \\   \Rightarrow \dfrac{1}{6} \times \left( {\dfrac{2}{5}{t^{\dfrac{5}{2}}} - 5 \times \dfrac{2}{3}{t^{\dfrac{3}{2}}} + C} \right) \\   \Rightarrow \dfrac{1}{{15}}{\left( {6x + 5} \right)^{\dfrac{5}{2}}} - \dfrac{5}{9}{\left( {6x + 5} \right)^{\dfrac{3}{2}}} + C \\   \Rightarrow \dfrac{1}{3}\left[ {\dfrac{{{{\left( {6x + 5} \right)}^{\dfrac{5}{2}}}}}{5} - \dfrac{{5{{\left( {6x + 5} \right)}^{\dfrac{3}{2}}}}}{3}} \right] + C \\  $

 Integrate the following term: $\int {\sin 14x\sin xdx} $.

 Use the formula $2\sin A\sin B = \cos \left( {A - B} \right) - \cos \left( {A + B} \right)$.

$  \int {\sin 14x\sin xdx}  \\   \Rightarrow \dfrac{1}{2}\int {2\sin 14\sin xdx}  \\  {\text{Apply }}2\sin A\operatorname{sinB}  = \cos \left( {A - B} \right) - \cos \left( {A + B} \right) \\   \Rightarrow \dfrac{1}{2}\int {\cos \left( {14x - x} \right)}  - \cos \left( {14x + x} \right)dx \\   \Rightarrow \dfrac{1}{2}\int {\cos 13xdx - \dfrac{1}{2}\int {\cos 15xdx} }  \\   \Rightarrow \dfrac{1}{2}\left[ {\dfrac{{\sin 13x}}{{13}} - \dfrac{{\sin 15x}}{{15}}} \right] + C \\  $

 Integrate the following term: ${\cot ^{ - 1}}\left( {\dfrac{{\sin 7x}}{{1 - \cos 7x}}} \right)dx$.

 Use $\sin 2A = 2\sin A\cos A$ and $1 - \cos 2A = {\sin ^2}A$ to find the integration.

$  \int {{{\cot }^{ - 1}}\left( {\dfrac{{\sin 7x}}{{1 - \cos 7x}}} \right)dx}  \\   \Rightarrow \int {{{\cot }^{ - 1}}\left( {\dfrac{{2\sin \dfrac{7}{2}x\cos \dfrac{7}{2}x}}{{2{{\sin }^2}\dfrac{7}{2}x}}} \right)dx}  \\   \Rightarrow \int {{{\cot }^{ - 1}}\left( {\cot \dfrac{7}{2}x} \right)dx}  \\   \Rightarrow \int {\dfrac{7}{2}x} dx = \dfrac{7}{4}{x^2} + C \\  $

 Find the integration of the following term: ${\tan ^{ - 1}}\left( {\sec 7x + \tan 7x} \right)dx$ .

 First simplify then term and then use $\dfrac{{\sin \left( {\dfrac{\pi }{2} + x} \right)}}{{\cos \left( {\dfrac{\pi }{2} + x} \right)}} = \tan \left( {\dfrac{\pi }{2} + x} \right)$ and $\sin x =  - \cos \left( {\dfrac{\pi }{2} + x} \right)$.

$  \int {{{\tan }^{ - 1}}\left( {\dfrac{1}{{\cos 7x}} + \dfrac{{\sin 7x}}{{\cos 7x}}} \right)dx}  \\   \Rightarrow \int {{{\tan }^{ - 1}}\left( {\dfrac{{1 + \sin 7x}}{{\cos 7x}}} \right)dx}  \\   \Rightarrow \int {{{\tan }^{ - 1}}\left[ {\dfrac{{1 - \cos \left( {\dfrac{\pi }{2} + 7x} \right)}}{{\sin \left( {\dfrac{\pi }{2} + 7x} \right)}}} \right]dx}  \\   \Rightarrow \int {{{\tan }^{ - 1}}\left[ {\dfrac{{2{{\sin }^2}\left( {\dfrac{\pi }{4} + \dfrac{7}{2}x} \right)}}{{2\sin \left( {\dfrac{\pi }{4} + \dfrac{7}{2}x} \right)\cos \left( {\dfrac{\pi }{4} + \dfrac{7}{2}x} \right)}}} \right]dx}  \\   \Rightarrow \int {{{\tan }^{ - 1}}\tan \left( {\dfrac{\pi }{4} + \dfrac{7}{2}x} \right)dx}  \\   \Rightarrow \dfrac{\pi }{4}x + \dfrac{7}{4}{x^2} + C \\  $