Advertisements
Advertisements
Question
Find `int (sin^2 x - cos^2x)/(sin x cos x) dx`
Advertisements
Solution
`int (sin^2 x - cos^2x)/(sin x cos x) dx`
= `int (sin^2 x)/(sin x cos x) dx - int (cos^2 x)/(sin x cos x) dx`
= `int tan x dx - int cot x dx`
= log |sec x| - log |sin x| + C
APPEARS IN
RELATED QUESTIONS
Evaluate : `intsin(x-a)/sin(x+a)dx`
Find the integrals of the function:
sin3 x cos3 x
Find the integrals of the function:
sin x sin 2x sin 3x
Find the integrals of the function:
`(1-cosx)/(1 + cos x)`
Find the integrals of the function:
`cos x/(1 + cos x)`
Find the integrals of the function:
`(sin^2 x)/(1 + cos x)`
Find the integrals of the function:
`(cos 2x - cos 2 alpha)/(cos x - cos alpha)`
Find the integrals of the function:
tan4x
Find the integrals of the function:
sin−1 (cos x)
Find `int dx/(x^2 + 4x + 8)`
Find `int (2x)/((x^2 + 1)(x^4 + 4))`dx
Find `int((3 sin x - 2) cos x)/(13 - cos^2 x- 7 sin x) dx`
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_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Find the area of the triangle whose vertices are (-1, 1), (0, 5) and (3, 2), using integration.
Find:
`int"dx"/sqrt(5-4"x" - 2"x"^2)`
Find: `int sec^2 x /sqrt(tan^2 x+4) dx.`
The value of the integral `int_(1/3)^1 (x - x^3)^(1/3)/x^4 dx` is
Which identity is used for an even power of cosine?
Which identity is used for the product of sine and cosine?
Which identity is used for \(\sin^3 x\)?
Using \(\cos^2 x = \frac{1+\cos 2x}{2}\), which expression is equivalent to \(\int \cos^2 x\,dx\)?
What is \(\int \sin 2x\cos 3x\,dx\)?
Which procedure is recommended when an identity can simplify a complicated trigonometric expression?
