English

∫cos7x dx - Mathematics and Statistics

Advertisements
Advertisements

Question

`int cos^7 x  "d"x`

Sum
Advertisements

Solution

Let I = `int cos^7 x  "d"x`

= `int(cos^2x)^3*cosx  "d"x`

= `int (1 - sin^2x)^3* cosx  "d"x`

Put sin x = t

∴ cos x dx = dt

∴ I = `int (1 - "t"^2)^3 "dt"`

= `int (1 - 3"t"^2 + 3"t"^4 - "t"^6) "dt"`

= `int 1* "dt" - 3 int "t"^2  "dt" + 3 int "t"^4 "dt" - int "t"^6 "dt"`

= `"t" - 3 ("t"^3/3) + 3"t"^5/5) - "t"^7/7 + "c"`

∴ I = `sinx - sin^3x + 3/5 sin^5x - 1/7 sin^7x + "c"`

shaalaa.com
  Is there an error in this question or solution?
Chapter 2.3: Indefinite Integration - Short Answers I

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:

`(e^(2x) - 1)/(e^(2x) + 1)`


Integrate the functions:

`(e^(2x) -  e^(-2x))/(e^(2x) + e^(-2x))`


Integrate the functions:

`cos sqrt(x)/sqrtx`


Evaluate: `int_0^3 f(x)dx` where f(x) = `{(cos 2x, 0<= x <= pi/2),(3, pi/2 <= x <= 3) :}`


\[\int\sqrt{3 + 2x - x^2} \text{ dx}\]

\[\int\sqrt{x - x^2} dx\]

\[\int\sqrt{9 - x^2}\text{ dx}\]

Write a value of

\[\int \tan^6 x \sec^2 x \text{ dx }\] .

Write a value of\[\int\frac{1}{x \left( \log x \right)^n} \text { dx }\].


Write a value of\[\int\sqrt{9 + x^2} \text{ dx }\].


Integrate the following w.r.t. x:

`2x^3 - 5x + 3/x + 4/x^5`


Evaluate the following integrals: `int sin 4x cos 3x dx`


Integrate the following functions w.r.t. x : `(1)/(4x + 5x^-11)`


Integrate the following functions w.r.t. x : `(1)/(x(x^3 - 1)`


Evaluate the following : `int (1)/(4x^2 - 3).dx`


Evaluate the following : `int (1)/(7 + 2x^2).dx`


Evaluate the following : `int (1)/sqrt(11 - 4x^2).dx`


Evaluate the following : `int (1)/sqrt(3x^2 + 5x + 7).dx`


Evaluate the following integrals:

`int (7x + 3)/sqrt(3 + 2x - x^2).dx`


Choose the correct options from the given alternatives :

`2 int (cos^2x - sin^2x)/(cos^2x + sin^2x)*dx` =


`int logx/(log ex)^2*dx` = ______.


Integrate the following w.r.t.x: `(3x + 1)/sqrt(-2x^2 + x + 3)`


Evaluate the following.

`int "x" sqrt(1 + "x"^2)` dx


Evaluate the following.

`int ((3"e")^"2t" + 5)/(4"e"^"2t" - 5)`dt


Evaluate the following.

`int (2"e"^"x" + 5)/(2"e"^"x" + 1)`dx


Evaluate:

`int (5x^2 - 6x + 3)/(2x − 3)` dx


Evaluate: `int "x" * "e"^"2x"` dx


Evaluate: `int sqrt("x"^2 + 2"x" + 5)` dx


Evaluate: `int sqrt(x^2 - 8x + 7)` dx


`int ("e"^(3x))/("e"^(3x) + 1)  "d"x`


`int(log(logx))/x  "d"x`


`int(5x + 2)/(3x - 4) dx` = ______


`int (cos x)/(1 - sin x) "dx" =` ______.


If I = `int (sin2x)/(3x + 4cosx)^3 "d"x`, then I is equal to ______.


`int(3x + 1)/(2x^2 - 2x + 3)dx` equals ______.


Evaluate `int 1/("x"("x" - 1)) "dx"`


`int dx/((x+2)(x^2 + 1))`    ...(given)

`1/(x^2 +1) dx = tan ^-1 + c`


Evaluate `int (1)/(x(x - 1))dx`


Evaluate.

`int (5x^2 - 6x + 3)/(2x - 3) dx`


Evaluate `int 1/(x(x-1))dx`


Evaluate the following.

`intx sqrt(1 +x^2)  dx`


`int (cos4x)/(sin2x + cos2x)dx` = ______.


Evaluate the following:

`int x^3/(sqrt(1+x^4))dx`


Evaluate the following.

`intx^3/sqrt(1+x^4)dx`


Evaluate the following.

`int "x"^3/sqrt(1 + "x"^4)` dx


Evaluate `int1/(x(x - 1))dx`


Evaluate `int (5x^2 - 6x + 3)/(2x - 3) dx`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×