Advertisements
Advertisements
Question
Evaluate: `int_0^(π/4) sec^4 x dx`
Advertisements
Solution
Let I = `int_0^(π/4) sec^4 x dx`
= `int_0^(π/4) sec^2 x . sec^2 x dx`
= `int_0^(π/4) (1 + tan^2x) sec^2 x dx`
Put tan x = t
∴ sec2 x dx = dt
Also, if x = 0, then t = 0
And if x = `π/4`, then t = 1
∴ I = `int_0^1 (1 + t^2) dt`
= `int_0^1 dt + int_0^1 t^2 dt`
= `[t]_0^1 + [t^3/3]_0^1`
= `1 + 1/3`
= `4/3`
APPEARS IN
RELATED QUESTIONS
Evaluate: `int_0^1 (x^2 - 2)/(x^2 + 1).dx`
Evaluate: `int_0^(pi/2) x sin x.dx`
Evaluate: `int_0^π sin^3x (1 + 2cosx)(1 + cosx)^2.dx`
Evaluate the following:
`int_0^a (1)/(x + sqrt(a^2 - x^2)).dx`
`int_0^(x/4) sqrt(1 + sin 2x) "d"x` =
`int_0^(pi/2) log(tanx) "d"x` =
Evaluate: `int_(- pi/4)^(pi/4) x^3 sin^4x "d"x`
Evaluate: `int_0^1 1/(1 + x^2) "d"x`
Evaluate: `int_0^(pi/4) sec^2 x "d"x`
Evaluate: `int_0^1 |x| "d"x`
Evaluate: `int_0^1 1/sqrt(1 - x^2) "d"x`
Evaluate: `int_(pi/6)^(pi/3) sin^2 x "d"x`
Evaluate: `int_0^(pi/2) sqrt(1 - cos 4x) "d"x`
Evaluate: `int_0^pi cos^2 x "d"x`
Evaluate: `int_0^(pi/4) cosx/(4 - sin^2 x) "d"x`
Evaluate: `int_(-1)^1 |5x - 3| "d"x`
Evaluate: `int_(-4)^2 1/(x^2 + 4x + 13) "d"x`
Evaluate: `int_0^(pi/4) sec^4x "d"x`
Evaluate: `int_0^(pi/2) 1/(5 + 4cos x) "d"x`
Evaluate: `int_0^1 "t"^2 sqrt(1 - "t") "dt"`
Evaluate: `int_(-1)^1 (1 + x^2)/(9 - x^2) "d"x`
Evaluate: `int_0^1 (1/(1 + x^2)) sin^-1 ((2x)/(1 + x^2)) "d"x`
Evaluate: `int_0^(pi/4) log(1 + tanx) "d"x`
Evaluate: `int_0^pi 1/(3 + 2sinx + cosx) "d"x`
If `int_2^e [1/logx - 1/(logx)^2].dx = a + b/log2`, then ______.
Evaluate:
`int_(π/4)^(π/2) cot^2x dx`.
Evaluate: `int_0^1 tan^-1(x/sqrt(1 - x^2))dx`.
Evaluate:
`int_(-π/2)^(π/2) |sinx|dx`
Evaluate `int_(π/6)^(π/3) cos^2x dx`
Evaluate:
`int_-4^5 |x + 3|dx`
The value of `int_2^(π/2) sin^3x dx` = ______.
Evaluate:
`int_0^(π/2) (sin 2x)/(1 + sin^4x)dx`
Find the value of ‘a’ if `int_2^a (x + 1)dx = 7/2`
Evaluate:
`int_0^(π/2) sinx/(1 + cosx)^3 dx`
Prove that: `int_0^1 logx/sqrt(1 - x^2)dx = π/2 log(1/2)`
