Test Index

Indefinite Integration

Section: Mathematics

© examsnet.com

Question : 30 of 89

Marks: +1, -0

If

\int x^2 \cos^2 x dx = \;\frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + c

f(1) + g(2) + h\left( \;\frac{1}{2} \right) =

[AP EAPCET 21 May 2025 S2]

0
2
1
-1

Solution:

Given,

\int x^2 \cos^2 x dx = \;\frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + C

LHS

I\; = \int x^2 \cos^2 x dx = \int x^2 \left( \frac{1+\cos 2x}{2} \right) dx

I\; = \;\frac{1}{2} \int (x^2 + x^2 \cos 2x) dx

\; = \;\frac{1}{2} \int x^2 dx + \;\frac{1}{2} \int x^2 \cos 2x dx

\; = \;\frac{1}{2} \cdot \;\frac{x^3}{3} + \;\frac{1}{2} I_1

Where

I_1 = \int x^2 \cos^2 x dx

\;I_1 = \;\frac{x^2}{2} \sin 2x + \;\frac{1}{2} x \cos 2x - \;\frac{1}{4} \sin 2x + C

\; \Rightarrow I = \;\frac{x^3}{6} + \;\frac{1}{2} \left\{ \sin 2x \left( \;\frac{x^2}{2} - \;\frac{1}{4} \right) + \;\frac{x}{2} \cos 2x \right\}

\; = \;\frac{x^3}{6} + \left( \;\frac{x^2}{4} - \;\frac{1}{8} \right) \sin 2x + \;\frac{x}{4} \cos 2x

\; = \;\frac{1}{6} f(x) + g(x) \sin 2x + h(x) \cos 2x + C

\; \because f(x)=x^3, g(x)=\;\frac{x^2}{4} - \;\frac{1}{8} \; \text{ and } \; h(x)=\;\frac{x}{4}

\; \Rightarrow f(1)=1, g(2)=\;\frac{7}{8} h\left( \;\frac{1}{2} \right) = \;\frac{1}{8}

\; \therefore f(1)+g(2)+h\left( \;\frac{1}{2} \right) = 1+\;\frac{7}{8}+\;\frac{1}{8}=2

© examsnet.com

Go to Question:

Prev Question

Next Question