JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 3)
Let n $$ \ge $$ 2 be a natural number and $$0 < \theta < {\pi \over 2}.$$ Then $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$ is equal to - (where C is a constant of integration)
$${n \over {{n^2} - 1}}{\left( {1 + {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n + 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
$${n \over {{n^2} - 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
$${n \over {{n^2} + 1}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{{n + 1} \over n}}} + C$$
Explanation
$$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
$$ = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
Put $$1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t$$
So $${{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt$$
Now $${1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt} $$
$$ = {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C$$
$$ = {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C$$
$$ = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
Put $$1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t$$
So $${{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt$$
Now $${1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt} $$
$$ = {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C$$
$$ = {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C$$
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