Advertisements
Advertisements
Question
Evaluate:
`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`
Solution
`int_(-pi/4)^(pi/4) (1)/(1 - sinx)*dx`
= `int_(-pi/4)^(pi/4) (1)/(1 - sinx)*(1 + sinx)/(1 + sinx)*dx`
= `int_(-pi/4)^(pi/4)(1 + sinx)/(1 - sin^2x)*dx`
= `int_(-pi/4)^(pi/4)(1 + sinx)/(cos^2x)*dx`
= `int_(-pi/4)^(pi/4) 1/(cos^2x) dx + int_(-pi/4)^(pi/4) (sinx)/(cos^2x)*dx`
= `int_(-pi/4)^(pi/4)sec^2x dx + int_(-pi/4)^(pi/4) 1/cosx*sinx/cosxdx`
= `[tanx]_(-pi/4)^(pi/4) + int_(-pi/4)^(pi/4) secx tanx*dx`
= `[tan pi/4 - tan (-pi/4)] + [secx]_(-pi/4)^(pi/4)`
= `[1 + 1] + [sec pi/4 - sec ((-pi)/4)]`
= `2 + [sqrt2 - sqrt2]`
= 2 + 0
I = 2
RELATED QUESTIONS
Prove that:
`{:(int_(-a)^a f(x) dx = 2 int_0^a f(x) dx",", "If" f(x) "is an even function"),( = 0",", "if" f(x) "is an odd function"):}`
Evaluate : `int_0^(pi/4) sin 4x sin 3x *dx`
Evaluate : `int_0^(1/sqrt(2)) (sin^-1x)/(1 - x^2)^(3/2)*dx`
Evaluate : `int_0^(pi//4) (sin2x)/(sin^4x + cos^4x)*dx`
Evaluate:
`int_0^(pi/2) sqrt(cos x) sin^3x * dx`
Evaluate : `int_0^(pi/4) sec^4x*dx`
Evaluate:
`int_0^1 sqrt((1 - x)/(1 + x)) * dx`
Evaluate : `int _((1)/(sqrt(2)))^1 (e^(cos^-1x) sin^-1x)/(sqrt(1 - x^2))*dx`
Evaluate : `int_0^(pi/2) (sinx - cosx)/(1 + sinx cosx)*dx`
Evaluate the following : `int_(-1)^(1) (x^3 + 2)/sqrt(x^2 + 4)*dx`
Evaluate the following : `int_0^1 t^2 sqrt(1 - t)*dt`
Choose the correct option from the given alternatives :
If `dx/(sqrt(1 + x) - sqrt(x)) = k/(3)`, then k is equal to
Evaluate the following : `int_0^(pi/2) cosx/(3cosx + sinx)*dx`
Evaluate the following : `int_(pi/4)^(pi/2) (cos theta)/[cos theta/2 + sin theta/2]^3*d theta`
Evaluate the following : `int_0^1 (cos^-1 x^2)*dx`
Evaluate the following:
`int_0^pi x/(1 + sin^2x) * dx`
Evaluate the following : `int_(pi/5)^((3pi)/10) sinx/(sinx + cosx)*dx`
Evaluate the following : `int_0^(pi/2) [2 log (sinx) - log (sin 2x)]*dx`
Evaluate the following : `int_(-2)^(3) |x - 2|*dx`
Evaluate the following definite integral:
`int_4^9 (1)/sqrt(x)*dx`
Evaluate the following definite integrals: If `int_0^"a" (2x + 1)*dx` = 2, find the real value of a.
Evaluate the following integrals : `int_(-9)^9 x^3/(4 - x^2).dx`
Choose the correct alternative :
`int_2^3 x/(x^2 - 1)*dx` =
Choose the correct alternative :
`int_0^2 e^x*dx` =
Choose the correct alternative :
`int_(-7)^7 x^3/(x^2 + 7)*dx` =
Fill in the blank : `int_0^2 e^x*dx` = ________
Fill in the blank : If `int_0^"a" 3x^2*dx` = 8, then a = _______
Solve the following : `int_1^2 dx/(x(1 + logx)^2`
`int_0^1 sqrt((1 - x)/(1 + x)) "d"x` =
Prove that: `int_0^(2"a") "f"(x) "d"x = int_0^"a" "f"(x) "d"x + int_0^"a" "f"(2"a" - x) "d"x`
Choose the correct alternative:
`int_4^9 ("d"x)/sqrt(x)` =
Choose the correct alternative:
`int_(-2)^3 1/(x + 5) "d"x` =
If `int_0^"a" (2x + 1) "d"x` = 2, find a
Evaluate `int_2^3 x/((x + 2)(x + 3)) "d"x`
Evaluate `int_1^3 log x "d"x`
`int_((-pi)/8)^(pi/8) log ((2 - sin x)/(2 + sin x))` dx = ______.
Evaluate the following definite integrals:
`int _1^2 (3x) / ( (9 x^2 - 1)) * dx`
Evaluate the following definite integral:
`int_4^9 1/sqrtx dx`
Evaluate the following definite integral:
`int_1^2 (3x)/((9x^2 - 1))dx`
`int_0^4 1/sqrt(4x - x^2)dx` = ______.
Solve the following.
`int_1^3 x^2 log x dx`
Prove that `int_0^(2a) f(x)dx = int_0^a[f(x) + f(2a - x)]dx`
Evaluate the following definite intergral:
`int_1^3logxdx`
Solve the following.
`int_0^1 e^(x^2) x^3 dx`
Evaluate the following definite intergral:
`int_1^2 (3x)/ ((9x^2 -1)) dx`
Evaluate the following definite intergral:
`int_-2^3 1/(x+5).dx`
Evaluate the following integral:
`int_-9^9 x^3/(4-x^2) dx`
