Advertisements
Advertisements
Question
Find the integrals of the function:
sin 3x cos 4x
Advertisements
Solution
Let `I = int sin 3x cos 4 x dx`
`1/2 int [sin (3x + 4x) + sin (3x - 4x)] dx` ...[∵ 2 sin A cos B = sin (A + B) + sin (A - B)]
`= 1/2 int [sin 7 x + sin (-x)] dx`
`= 1/2 int (sin 7x - sin x) dx`
`= 1/2 ((-cos 7x)/7) - 1/2 (-cos x) + C`
`= -1/14 cos 7x + 1/2 cos x + C`
APPEARS IN
RELATED QUESTIONS
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
sin2 (2x + 5)
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
cos4 2x
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
tan3 2x sec 2x
Find the integrals of the function:
tan4x
Find the integrals of the function:
`(sin^3 x + cos^3 x)/(sin^2x cos^2 x)`
Find the integrals of the function:
`1/(sin xcos^3 x)`
Find the integrals of the function:
sin−1 (cos x)
Find the integrals of the function:
`1/(cos(x - a) cos(x - b))`
`int (sin^2x - cos^2 x)/(sin^2 x cos^2 x) dx` is equal to ______.
`int (e^x(1 +x))/cos^2(e^x x) dx` equals ______.
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Find `int dx/(x^2 + 4x + 8)`
Differentiate : \[\tan^{- 1} \left( \frac{1 + \cos x}{\sin x} \right)\] with respect to x .
Evaluate : \[\int\limits_0^\pi \frac{x \tan x}{\sec x \cdot cosec x}dx\] .
Find `int_ (sin "x" - cos "x" )/sqrt(1 + sin 2"x") d"x", 0 < "x" < π / 2 `
Find `int_ (log "x")^2 d"x"`
Find: `int_ (cos"x")/((1 + sin "x") (2+ sin"x")) "dx"`
Find: `intsqrt(1 - sin 2x) dx, pi/4 < x < pi/2`
`int "dx"/(sin^2x cos^2x)` is equal to ______.
Evaluate the following:
`int sqrt(1 + sinx)"d"x`
Evaluate the following:
`int (cosx - cos2x)/(1 - cosx) "d"x`
Evaluate the following:
`int sin^-1 sqrt(x/("a" + x)) "d"x` (Hint: Put x = a tan2θ)
`int (x + sinx)/(1 + cosx) "d"x` is equal to ______.
`int sinx/(3 + 4cos^2x) "d"x` = ______.
