Question

Evaluate : $$\int {\frac{1}{{1 + 3\,{{\sin }^2}x + 8\,{{\cos }^2}x}}} dx$$

A. $$\frac{1}{6}{\tan ^{ - 1}}\left( {2\,\tan \,x} \right) + C$$
B. $${\tan ^{ - 1}}\left( {2\,\tan \,x} \right) + C$$
C. $$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2\,\tan \,x}}{3}} \right) + C$$  
D. none of these
Answer :   $$\frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2\,\tan \,x}}{3}} \right) + C$$
Solution :
$$I = \int {\frac{1}{{1 + 3\,{{\sin }^2}x + 8\,{{\cos }^2}x}}} dx$$
Dividing the numerator and denominator by $${\cos ^2}x,$$  we ge
$$\eqalign{ & I = \int {\frac{{{{\sec }^2}x}}{{{{\sec }^2}x + 3\,{{\tan }^2}x + 8}}} dx \cr & \Rightarrow I = \int {\frac{{{{\sec }^2}x}}{{1 + {{\tan }^2}x + 3\,{{\tan }^2}x + 8}}dx} \cr & \Rightarrow I = \int {\frac{{{{\sec }^2}x}}{{4\,{{\tan }^2}x + 9}}} dx \cr & {\text{Putting}}\,\,\tan \,x = t \Rightarrow {\sec ^2}x\,dx = dt,\,{\text{we get}} \cr & \Rightarrow I = \int {\frac{{dt}}{{4{t^2} + 9}}} \cr & \Rightarrow I = \frac{1}{4}\int {\frac{{dt}}{{{t^2} + {{\left( {\frac{3}{2}} \right)}^2}}}} \cr & \Rightarrow I = \frac{1}{4} \times \frac{1}{{\frac{3}{2}}}{\tan ^{ - 1}}\left( {\frac{t}{{\frac{3}{2}}}} \right) + C \cr & \Rightarrow I = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2t}}{3}} \right) + C \cr & \Rightarrow I = \frac{1}{6}{\tan ^{ - 1}}\left( {\frac{{2\,\tan \,x}}{3}} \right) + C \cr} $$

Releted MCQ Question on
Calculus >> Indefinite Integration

Releted Question 1

The value of the integral $$\int {\frac{{{{\cos }^3}x + {{\cos }^5}x}}{{{{\sin }^2}x + {{\sin }^4}x}}dx} $$    is-

A. $$\sin \,x - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
B. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + c$$
C. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} - 6\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
D. $$\sin \,x - 2{\left( {\sin \,x} \right)^{ - 1}} + 5\,{\tan ^{ - 1}}\left( {\sin \,x} \right) + c$$
View Answer
Releted Question 2

If $$\int_{\sin \,x}^1 {{t^2}f\left( t \right)dt = 1 - \sin \,x} ,$$      then $$f\left( {\frac{1}{{\sqrt 3 }}} \right)$$   is-

A. $$\frac{1}{3}$$
B. $${\frac{1}{{\sqrt 3 }}}$$
C. $$3$$
D. $$\sqrt 3 $$
View Answer
Releted Question 3

Solve this $$\int {\frac{{{x^2} - 1}}{{{x^3}\sqrt {2{x^4} - 2{x^2} + 1} }}dx} = ?$$

A. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^2}}} + C$$
B. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{{x^3}}} + C$$
C. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{x} + C$$
D. $$\frac{{\sqrt {2{x^4} - 2{x^2} + 1} }}{{2{x^2}}} + C$$
View Answer
Releted Question 4

Let $$I = \int {\frac{{{e^x}}}{{{e^{4x}} + {e^{2x}} + 1}}dx,\,J = \int {\frac{{{e^{ - \,x}}}}{{{e^{ - \,4x}} + {e^{ - \,2x}} + 1}}dx.} } $$
Then for an arbitrary constant $$C,$$ the value of $$J-I$$  equals-

A. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} - {e^{2x}} + 1}}{{{e^{4x}} + {e^{2x}} + 1}}} \right) + C$$
B. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} + {e^x} + 1}}{{{e^{2x}} - {e^x} + 1}}} \right) + C$$
C. $$\frac{1}{2}\log \left( {\frac{{{e^{2x}} - {e^x} + 1}}{{{e^{2x}} + {e^x} + 1}}} \right) + C$$
D. $$\frac{1}{2}\log \left( {\frac{{{e^{4x}} + {e^{2x}} + 1}}{{{e^{4x}} - {e^{2x}} + 1}}} \right) + C$$
View Answer

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Indefinite Integration

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