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1

AIEEE 2012

MCQ (Single Correct Answer)
If the $$\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,} $$ then $$a$$ is
equal to :
A
$$-1$$
B
$$-2$$
C
$$1$$
D
$$2$$

Explanation

$$\int {{{5\tan x} \over {\tan x - 2}}} dx$$

$$ = \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx$$

$$ = \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$

$$ = \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}} $$

$$ = \int {\left( {{{4\sin x + \sin x + 2\cos x - 2\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$

$$ = \int {{{\left( {\sin x - 2\cos x} \right) + \left( {4\sin x + 2\cos x} \right)} \over {\sin x - 2\cos x}}} \,dx$$

$$ = \int {{{\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \over {\left( {\sin x - 2\cos x} \right)}}} \,dx$$

$$ = \int {{{\sin x - 2\cos x} \over {\sin x - 2\cos x}}dx + 2\int {\left( {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}} \right)} } \,dx$$

$$ = \int {dx + 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos \,x}}} } \,dx$$

$$ = {I_1} + {I_2}$$ where $${I_1} = \int {dx} $$ and

$${I_2} = 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}\,dx} $$

Put $$\sin x - 2\cos x = t$$

$$ \Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt$$

$$\therefore$$ $${I_2} = 2\int {{{dt} \over t}} = 2\ln \,t + C$$

$$ = 2\,\ln \left( {\sin \,x - 2\cos \,x} \right) + C$$

Hence, $${I_1} + {I_2}$$

$$ = \int {dx + 2\ln \left( {\sin x - 2\cos x} \right) + c} $$

$$ = x + 2\ln \left| {\left( {\sin x - 2\cos x} \right)} \right| + k$$

$$ \Rightarrow a = 2$$
2

AIEEE 2008

MCQ (Single Correct Answer)
The value of $$\sqrt 2 \int {{{\sin xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$ is
A
$$\,x + \log \,\left| {\,\cos \left( {x - {\pi \over 4}} \right)\,} \right| + c$$
B
$$\,x - \log \,\left| {\,\sin \left( {x - {\pi \over 4}} \right)\,} \right| + c$$
C
$$\,x + \log \,\left| {\,\sin \left( {x - {\pi \over 4}} \right)\,} \right| + c$$
D
$$\,x - \log \,\left| {\,\cos \left( {x - {\pi \over 4}} \right)\,} \right| + c$$

Explanation

Let $$I = \sqrt 2 \int {{{\sin \,xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$

Put $$x - {\pi \over 4} = t$$

$$ \Rightarrow dx = dt$$

$$ \Rightarrow I = \sqrt 2 \int {{{\sin \left( {t + {\pi \over 4}} \right)} \over {\sin \,t}}} dt$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = {{\sqrt 2 } \over {\sqrt 2 }}\int {\left( {{{\sin t + \cos t} \over {\sin t}}} \right)} \,\,dt$$

$$ \Rightarrow I = \int {\left( {1 + \cot \,t} \right)} dt$$

$$\,\,\,\,\,\,\,\,\,\,\,\, = t + \log \left| {\sin t} \right| + {c_1}$$

$$ = x - {\pi \over 4} + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + {c_1}$$

$$ = x + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + c$$

$$\left( \, \right.$$ where $${c = {c_1} - {\pi \over 4}}$$ $$\left. \, \right)$$
3

AIEEE 2007

MCQ (Single Correct Answer)
$$\int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$ equals
A
$$\log \,\tan \,\left( {{x \over 2} + {\pi \over {12}}} \right) + C$$
B
$$\log \,\tan \,\left( {{x \over 2} - {\pi \over {12}}} \right) + C$$
C
$$\,{1 \over 2}\,\log \,\tan \,\left( {{x \over 2} + {\pi \over {12}}} \right) + C$$
D
$$\,{1 \over 2}\,\log \,\tan \,\left( {{x \over 2} - {\pi \over {12}}} \right) + C$$

Explanation

$$I = \int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$

$$ \Rightarrow I = \int {{{dx} \over {2\left[ {{1 \over 2}\cos x + {{\sqrt 3 } \over 2}\sin x} \right]}}} $$

$$ = {1 \over 2}\int {{{dx} \over {\left[ {\sin {\pi \over 6}\cos x + \cos {\pi \over 6}\sin x} \right]}}} $$

$$ = {1 \over 2}.\int {{{dx} \over {\sin \left( {x + {\pi \over 6}} \right)}}} $$

$$ \Rightarrow I = {1 \over 2}.\int {\cos ec\left( {x + {\pi \over 6}} \right)dx} $$

But we know that

$$\int {\cos ec\,x\,dx} = \log \left| {\left( {\tan x/2} \right)} \right| + C$$

$$\therefore$$ $$I = {1 \over 2}.\log \,\tan \left( {{x \over 2} + {\pi \over {12}}} \right) + C$$
4

AIEEE 2005

MCQ (Single Correct Answer)
$$\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} $$ is equal to
A
$${{\log x} \over {{{\left( {\log x} \right)}^2} + 1}} + C$$
B
$${x \over {{x^2} + 1}} + C$$
C
$${{x{e^x}} \over {1 + {x^2}}} + C$$
D
$${x \over {{{\left( {\log x} \right)}^2} + 1}} + C$$

Explanation

$$\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx$$

$$ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} $$

$$ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} \right)}^2}} \right)}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx$$

$$ = \int {\left[ {{{{e^t}} \over {1 + {t^2}}} - {{2t\,{e^t}} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt$$

put $$\log x = t \Rightarrow dx = {e^t}\,dt$$

$$ = \int {{e^t}} \left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt$$

$$\left[ \, \right.$$ which is of the form

$$\left. {\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)dx} \right)} } \right]$$

$$ = {{{e^t}} \over {1 + {t^2}}} + c = {x \over {1 + {{\left( {\log x} \right)}^2}}} + c$$

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