Advertisements
Advertisements
Question
\[\int\limits_0^{\pi/4} \tan^4 x dx\]
Advertisements
Solution
\[\int_0^\frac{\pi}{4} \tan^4 x d x\]
\[ = \int_0^\frac{\pi}{4} \tan^2 x\left( se c^2 x - 1 \right) d x\]
\[ = \int_0^\frac{\pi}{4} \tan^2 x se c^2 x dx - \int_0^\frac{\pi}{4} \tan^2 x dx\]
\[ = \left[ \frac{\tan^3 x}{3} \right]_0^\frac{\pi}{4} - \left[ \tan x - x \right]_0^\frac{\pi}{4} \]
\[ = \frac{1}{3} - 1 + \frac{\pi}{4}\]
\[ = \frac{\pi}{4} - \frac{2}{3}\]
APPEARS IN
RELATED QUESTIONS
Evaluate the following definite integrals:
Evaluate each of the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
If \[\int_0^a \frac{1}{4 + x^2}dx = \frac{\pi}{8}\] , find the value of a.
Evaluate : \[\int\limits_0^\pi/4 \frac{\sin x + \cos x}{16 + 9 \sin 2x}dx\] .
Evaluate : \[\int\frac{dx}{\sin^2 x \cos^2 x}\] .
\[\int\limits_0^\pi \sin^3 x\left( 1 + 2 \cos x \right) \left( 1 + \cos x \right)^2 dx\]
\[\int\limits_0^{\pi/2} x^2 \cos 2x dx\]
\[\int\limits_0^1 x \left( \tan^{- 1} x \right)^2 dx\]
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
\[\int\limits_0^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} dx\]
\[\int\limits_0^1 \cot^{- 1} \left( 1 - x + x^2 \right) dx\]
\[\int\limits_0^4 x dx\]
\[\int\limits_{- 1}^1 e^{2x} dx\]
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_1^2 (x - 1)/x^2 "d"x`
Evaluate the following using properties of definite integral:
`int_0^1 x/((1 - x)^(3/4)) "d"x`
Evaluate the following:
`Γ (9/2)`
Evaluate `int "dx"/sqrt((x - alpha)(beta - x)), beta > alpha`
Verify the following:
`int (x - 1)/(2x + 3) "d"x = x - log |(2x + 3)^2| + "C"`
What is the result of a definite integral?
If velocity changes with time, what does a definite integral give over a time interval?
