JEE MAIN - Mathematics (2019 - 9th April Morning Slot - No. 18)
The integral $$\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}} $$ is equal to
(Hence C is a constant of integration)
-3/4 tan - 4 / 3 x + C
3tan–1/3x + C
–3cot–1/3x+ C
- 3tan–1/3x + C
Explanation
$$\int {{{\sec }^{{2 \over 3}}}} x\cos e{c^{{4 \over 3}}}xdx$$
= $$\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
= $$\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
Let cot x = t3
$$ \Rightarrow $$ - cosec2x dx = 3t2dt
= $$ - 3\int {{{{t^2}dt} \over {{t^2}}}} $$
= -3t + C
= -3$${{{\cot }^{{1 \over 3}}}x}$$ + C
= -3$${{{\tan }^{ - {1 \over 3}}}x}$$ + C
= $$\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
= $$\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
Let cot x = t3
$$ \Rightarrow $$ - cosec2x dx = 3t2dt
= $$ - 3\int {{{{t^2}dt} \over {{t^2}}}} $$
= -3t + C
= -3$${{{\cot }^{{1 \over 3}}}x}$$ + C
= -3$${{{\tan }^{ - {1 \over 3}}}x}$$ + C
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