JEE MAIN - Mathematics (2017 (Offline) - No. 15)
Let $${I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right).$$
If $${I_4} + {I_6}$$ = $$a{\tan ^5}x + b{x^5} + C$$, where C is a constant of integration,
then the ordered pair $$\left( {a,b} \right)$$ is equal to
If $${I_4} + {I_6}$$ = $$a{\tan ^5}x + b{x^5} + C$$, where C is a constant of integration,
then the ordered pair $$\left( {a,b} \right)$$ is equal to
$$\left( {{1 \over 5},0} \right)$$
$$\left( {{1 \over 5}, - 1} \right)$$
$$\left( { - {1 \over 5},0} \right)$$
$$\left( { - {1 \over 5},1} \right)$$
Explanation
Given,
In = $$\int {{{\tan }^n}x\,dx,\,\,\,n > 1} $$
$$\therefore\,\,\,$$ I4 = $$\int {{{\tan }^4}x\,dx} $$
and I6 = $$\int {{{\tan }^6}} x\,dx$$
$$\therefore\,\,\,$$ I = I4 + I6
= $$\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)} dx$$
= $$\int {{{\tan }^4}} x\left( {1 + {{\tan }^2}x} \right)dx$$
= $$\int {{{\tan }^4}} x.{\sec ^2}x\,dx$$
Let, tanx = t
$$ \Rightarrow $$$$\,\,\,$$ sec2x dx = dt
$$\therefore\,\,\,$$ I = $$\int {{t^4}\,dt} $$
= $${1 \over 5}$$ t5 + C
= $${1 \over 5}$$ tan5x + C
$$\therefore\,\,\,$$ By comparing with the question, we get
A = $${1 \over 5}$$, B = 0
In = $$\int {{{\tan }^n}x\,dx,\,\,\,n > 1} $$
$$\therefore\,\,\,$$ I4 = $$\int {{{\tan }^4}x\,dx} $$
and I6 = $$\int {{{\tan }^6}} x\,dx$$
$$\therefore\,\,\,$$ I = I4 + I6
= $$\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)} dx$$
= $$\int {{{\tan }^4}} x\left( {1 + {{\tan }^2}x} \right)dx$$
= $$\int {{{\tan }^4}} x.{\sec ^2}x\,dx$$
Let, tanx = t
$$ \Rightarrow $$$$\,\,\,$$ sec2x dx = dt
$$\therefore\,\,\,$$ I = $$\int {{t^4}\,dt} $$
= $${1 \over 5}$$ t5 + C
= $${1 \over 5}$$ tan5x + C
$$\therefore\,\,\,$$ By comparing with the question, we get
A = $${1 \over 5}$$, B = 0
Comments (0)
