Advertisements
Advertisements
Question
Evaluate each of the following integral:
Advertisements
Solution
\[\int_0^\frac{\pi}{4} \tan xdx\]
\[ = \left.{\log\sec\ x}\right|_0^\frac{\pi}{4} \]
\[ = \log\sec\frac{\pi}{4} - \log\sec0\]
\[ = \log\sqrt{2} - \log1\]
\[ = \log 2^\frac{1}{2} - 0\]
\[ = \frac{1}{2}\log2\]
APPEARS IN
RELATED QUESTIONS
If `f` is an integrable function such that f(2a − x) = f(x), then prove that
Given that \[\int\limits_0^\infty \frac{x^2}{\left( x^2 + a^2 \right)\left( x^2 + b^2 \right)\left( x^2 + c^2 \right)} dx = \frac{\pi}{2\left( a + b \right)\left( b + c \right)\left( c + a \right)},\] the value of \[\int\limits_0^\infty \frac{dx}{\left( x^2 + 4 \right)\left( x^2 + 9 \right)},\]
Evaluate: \[\int\limits_{- \pi/2}^{\pi/2} \frac{\cos x}{1 + e^x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^{\pi/2} \frac{\sin x}{\sqrt{1 + \cos x}} dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_0^{\pi/2} \frac{dx}{4 \cos x + 2 \sin x}dx\]
\[\int\limits_0^2 \left( 2 x^2 + 3 \right) dx\]
\[\int\limits_1^4 \left( x^2 + x \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Using second fundamental theorem, evaluate the following:
`int_(-1)^1 (2x + 3)/(x^2 + 3x + 7) "d"x`
Evaluate the following using properties of definite integral:
`int_(-1)^1 log ((2 - x)/(2 + x)) "d"x`
Choose the correct alternative:
`int_0^oo "e"^(-2x) "d"x` is
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Find: `int logx/(1 + log x)^2 dx`
Evaluate: `int_(-1)^2 |x^3 - 3x^2 + 2x|dx`
