IIT JEE Main Maths -Unit 8 -Indefinite Integrations Methods- Exam Style Questions- New Syllabus
▶ Answer/Explanation
Ans. (3)
\(
\therefore \frac{d}{dx} \left( \frac{x \sin^{-1} x}{\sqrt{1 – x^2}} \right)
= \frac{\sin^{-1} x}{(1 – x^2)^{3/2}} + \frac{x}{1 – x^2}
\)
\(
\Rightarrow \int e^x \left( \frac{x \sin^{-1} x}{\sqrt{1 – x^2}}
+ \frac{\sin^{-1} x}{(1 – x^2)^{3/2}} + \frac{x}{1 – x^2} \right) dx
\)
\(
= e^x \cdot \frac{x \sin^{-1} x}{\sqrt{1 – x^2}} + c = g(x) + C
\)
\(
\text{Note : assuming } g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1 – x^2}}
\)
\(
g\left( \frac{1}{2} \right) = \frac{e^{1/2}}{2} \cdot \frac{\pi/6 \times 2}{\sqrt{3}}
= \frac{\pi}{6} \sqrt{\frac{e}{3}}
\)
Comment : In this question we will not get a unique function } g(x),
but in order to match the answer we will have to assume
\(g(x) = \frac{x e^x \sin^{-1} x}{\sqrt{1 – x^2}}.\)
▶️ Answer/Explanation
\(\text{Ans. } (2)\)
\(\text{Sol. } I(x) = \int \frac{dx}{(x – 11)^{\frac{11}{13}} (x + 15)^{\frac{15}{13}}}\)
\(\text{Put } \frac{x – 11}{x + 15} = t \Rightarrow \frac{26}{(x + 15)^2} dx = dt\)
\(\Rightarrow I(x) = \frac{1}{26} \int \frac{dt}{t^{\frac{11}{13}}} = \frac{1}{26} \cdot \frac{t^{\frac{2}{13}}}{\frac{2}{13}}\)
\(\therefore I(x) = \frac{1}{4} \left( \frac{x – 11}{x + 15} \right)^{\frac{2}{13}} + C\)
$I(37) – I(24) = \frac{1}{4} \left[ \left( \frac{26}{52} \right)^{\frac{2}{13}} – \left( \frac{13}{39} \right)^{\frac{2}{13}} \right]$
$= \frac{1}{4} \left( \frac{1}{2^{\frac{2}{13}}} – \frac{1}{3^{\frac{2}{13}}} \right)$
\(\therefore b = 4, \, c = 9\)
\(\Rightarrow 3(b + c) = 3(4 + 9) = 39\)