Advertisements
Advertisements
Question
Advertisements
Solution
Let I= \[\int\]sin3 x . cos x dx
⇒ cos x dx = dt
\[ = \frac{\sin^4 x}{4} + C \left( \because t = \sin x \right)\]
APPEARS IN
RELATED QUESTIONS
Evaluate : `int_0^pi(x)/(a^2cos^2x+b^2sin^2x)dx`
Find: `int(x+3)sqrt(3-4x-x^2dx)`
Integrate the functions:
`x/(e^(x^2))`
Integrate the functions:
`(e^(2x) - e^(-2x))/(e^(2x) + e^(-2x))`
Integrate the functions:
`cos sqrt(x)/sqrtx`
Integrate the functions:
`1/(1 - tan x)`
`(10x^9 + 10^x log_e 10)/(x^10 + 10^x) dx` equals:
Write a value of\[\int\frac{1}{1 + 2 e^x} \text{ dx }\].
Write a value of\[\int\frac{\left( \tan^{- 1} x \right)^3}{1 + x^2} dx\]
Write a value of\[\int a^x e^x \text{ dx }\]
Integrate the following w.r.t. x : `int x^2(1 - 2/x)^2 dx`
Integrate the following functions w.r.t. x : `(1)/(2 + 3tanx)`
Integrate the following functions w.r.t. x : `(sin6x)/(sin 10x sin 4x)`
Evaluate the following:
`int (1)/(25 - 9x^2)*dx`
Evaluate the following : `int (1)/(x^2 + 8x + 12).dx`
Evaluate the following integrals : `int (2x + 3)/(2x^2 + 3x - 1).dx`
Evaluate the following : `int (logx)2.dx`
Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`
Choose the correct alternative from the following.
The value of `int "dx"/sqrt"1 - x"` is
State whether the following statement is True or False.
The proper substitution for `int x(x^x)^x (2log x + 1) "d"x` is `(x^x)^x` = t
`int (log x)/(log ex)^2` dx = _________
`int x/(x + 2) "d"x`
`int (1 + x)/(x + "e"^(-x)) "d"x`
`int sin^-1 x`dx = ?
`int ((x + 1)(x + log x))^4/(3x) "dx" =`______.
`int ("d"x)/(sinx cosx + 2cos^2x)` = ______.
`int sqrt(x^2 - a^2)/x dx` = ______.
`int x/sqrt(1 - 2x^4) dx` = ______.
(where c is a constant of integration)
Evaluate `int (1+x+x^2/(2!))dx`
Evaluate.
`int (5x^2 - 6x + 3)/(2x - 3) dx`
Evaluate the following.
`int(1)/(x^2 + 4x - 5)dx`
Evaluate:
`int sin^3x cos^3x dx`
Evaluate the following.
`int1/(x^2 + 4x - 5) dx`
Evaluate the following.
`int1/(x^2+4x-5)dx`
Evaluate the following.
`intx^3/sqrt(1 + x^4) dx`
