Advertisements
Advertisements
Question
Solution
\[\int_0^\frac{\pi}{4} \left( a^2 \cos^2 x + b^2 \sin^2 x \right)dx\]
\[ = \int_0^\frac{\pi}{4} \left[ a^2 \left( \frac{1 + \cos2x}{2} \right) + b^2 \left( \frac{1 - \cos2x}{2} \right) \right]dx\]
\[ = \int_0^\frac{\pi}{4} \left[ \left( \frac{a^2 + b^2}{2} \right) + \left( \frac{a^2 - b^2}{2} \right)\cos2x \right]dx\]
\[ = \left( \frac{a^2 + b^2}{2} \right) \int_0^\frac{\pi}{4} dx + \left( \frac{a^2 - b^2}{2} \right) \int_0^\frac{\pi}{4} \cos2xdx\]
\[= \left.\left( \frac{a^2 + b^2}{2} \right) \times x\right|_0^\frac{\pi}{4} + \left.\left( \frac{a^2 - b^2}{2} \right) \times \frac{\sin2x}{2}\right|_0^\frac{\pi}{4} \]
\[ = \left( \frac{a^2 + b^2}{2} \right)\left( \frac{\pi}{4} - 0 \right) + \left( \frac{a^2 - b^2}{4} \right)\left( \sin\frac{\pi}{2} - \sin0 \right)\]
\[ = \left( \frac{a^2 + b^2}{2} \right)\frac{\pi}{4} + \left( \frac{a^2 - b^2}{4} \right)\left( 1 - 0 \right)\]
\[ = \left( a^2 + b^2 \right)\frac{\pi}{8} + \frac{1}{4}\left( a^2 - b^2 \right)\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate each of the following integral:
\[\int_a^b \frac{x^\frac{1}{n}}{x^\frac{1}{n} + \left( a + b - x \right)^\frac{1}{n}}dx, n \in N, n \geq 2\]
If f(2a − x) = −f(x), prove that
\[\int\limits_0^\infty \frac{1}{1 + e^x} dx\] equals
\[\int\limits_1^5 \frac{x}{\sqrt{2x - 1}} dx\]
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
\[\int\limits_0^\pi \frac{x}{1 + \cos \alpha \sin x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
Evaluate the following using properties of definite integral:
`int_(- pi/4)^(pi/4) x^3 cos^3 x "d"x`
Choose the correct alternative:
`int_0^1 (2x + 1) "d"x` is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
Verify the following:
`int (2x + 3)/(x^2 + 3x) "d"x = log|x^2 + 3x| + "C"`
