Integrals

Get insights from 25 questions on Integrals, answered by students, alumni, and experts. You may also ask and answer any question you like about Integrals

25
Questions
0
Discussions
0
Active Users
0
Followers

Have a question about Integrals? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Integrals Class 12th

If $\frac{}{^{} {(^{})}^{}} {(^{})}^{}$ $\int \frac{c o s ? x d x}{{s i n}^{3} ? x {(1 + {s i n}^{6} ? x)}^{2} 3} = f (x) {(1 + {s i n}^{6} ? x)}^{1 / λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ $(\frac{}{})$ is equal to:

7 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

sinx = t, cosxdx = dt
I = ∫dt/(t³(1+t?)^(2/3)) = ∫dt/(t?(1+1/t?)^(2/3))
Put 1+1/t? = r³ ⇒ -6/t? dt = 3r²dr
I = (-1/2)∫dr = -1/2 r + c = (-1/2)(1+1/sin?x)^(1/3) + c
f(x) = (-1/2)cosec²x(1+sin?x)^(1/3) and λ=3
⇒ f(π/3) = -2

Upvote

New answer posted•a year ago

Integrals Class 12th

If ∫(cosθ/(5+7sinθ-2cos²θ))dθ = A logₑ|B(θ)| + C, where C is a constant of integration, then B(θ)/A can be:

14 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

I = ∫ (cosθ / (2sin²θ + 7sinθ + 3) dθ
sinθ = t ⇒ cosθdθ = dt
= (1/2) ∫ (1 / (t² + (7/2)t + 3/2) dt
= (1/2) ∫ (1 / (t+7/4)² - (5/4)²) dt
= (1/5) ln | (2t+1)/ (t+3)| + c
= (1/5) ln | (2sinθ+1)/ (sinθ+3)| + C
So A = 1/5
B (θ) = 5 (2sinθ+1)/ (sinθ+3)

Upvote

New answer posted•a year ago

Integrals Class 12th

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 a n d l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2} a n d l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} i s :$

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$? l_{1} a n d l_{2}$ are perpendicular, so

$3 \times 1 + (- 2) (\frac{α}{2}) + 0 \times 2 = 0$

a = 3

Now angle between $l_{2} a n d l_{3}$ ,

$c o s θ = \frac{1 (- 3) + \frac{α}{2} (- 2) + 2 (4)}{\sqrt[]{1 + \frac{α^{2}}{4} + 4} . \sqrt[]{9 + 4 + 16}}$

$\Rightarrow c o s θ = \frac{\frac{2}{29}}{2} \Rightarrow θ = c o s^{- 1} (\frac{4}{29}) = s e c^{- 1} (\frac{29}{4})$

Upvote

New answer posted•a year ago

Integrals Class 12th

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$ is equal to

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$

let $c o s^{- 1} x = \frac{π}{4} + θ$

= $\underset{θ \to \infty}{l i m} \frac{2 s i n θ}{- 2 t a n θ} (1 - t a n θ) = -$ $1$

Upvote

Integrals

Questions

Discussions

Active Users

Followers

If ∫ c o s ? x d x s i n 3 ? x 1 + s i n 6 ? x 2 3 = f ( x ) 1 + s i n 6 ? x 1 / λ + c where c is a constant of integration, then λ f π 3 is equal to:

If ∫(cosθ/(5+7sinθ-2cos²θ))dθ = A logₑ|B(θ)| + C, where C is a constant of integration, then B(θ)/A can be:

If the two lines l1:x−23=y+1−2,z=2 and l2:x−11=2y+3α=z+52 are perpendicular, then an angle between the lines l2and l3:1−x3=2y−1−4=z4 is:

limx→12sin(cos−1x)−x1−tan(cos−1x) is equal to

If $\frac{}{^{} {(^{})}^{}} {(^{})}^{}$ $\int \frac{c o s ? x d x}{{s i n}^{3} ? x {(1 + {s i n}^{6} ? x)}^{2} 3} = f (x) {(1 + {s i n}^{6} ? x)}^{1 / λ} + c$ where $c$ is a constant of integration, then $λ f (\frac{π}{3})$ $(\frac{}{})$ is equal to:

If the two lines $l_{1} : \frac{x - 2}{3} = \frac{y + 1}{- 2}, z = 2 a n d l_{2} : \frac{x - 1}{1} = \frac{2 y + 3}{α} = \frac{z + 5}{2}$ are perpendicular, then an angle between the lines $l_{2} a n d l_{3} : \frac{1 - x}{3} = \frac{2 y - 1}{- 4} = \frac{z}{4} i s :$

$\underset{x \to \frac{1}{\sqrt[]{2}}}{l i m} \frac{s i n (c o s^{- 1} x) - x}{1 - t a n (c o s^{- 1} x)}$ is equal to