-sin-cos- fx tan- dx is where -n-2- n - I

Let x/cosθ = t ⇒ dx = (cosθ) dt ∫_sinθ^cosθ f(x tanθ) dx =∫^1_tanθ f(t sinθ) cosθ dt = cosθ∫_1^tanθ f(t sinθ)dt Put t=x then we get, nbsp; = nbsp; cosθ ...

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Byju's Answer

Standard XIII

Mathematics

Integration by Substitution

∫sinθcosθ fx ...

Question

$\int_{sin θ}^{cos θ} f (x tan θ) d x$ is
$(w h e r e θ \neq \frac{n π}{2}, n \in I)$

- cos θ \int_{1}^{tan θ} f (x sin θ) d x

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- tan θ \int_{sin θ}^{cos θ} f (x) d x

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sin θ \int_{0}^{tan θ} f (x cos θ) d x

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cot θ \int_{sin θ}^{sin θ tan θ} f (x) d x

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Solution

The correct option is A $- cos θ \int_{1}^{tan θ} f (x sin θ) d x$
$L e t \frac{x}{cos θ} = t \Rightarrow d x = (cos θ) d t$
$\int_{sin θ}^{cos θ} f (x tan θ) d x$
$= \int_{tan θ}^{1} f (t sin θ) cos θ d t = - cos θ \int_{1}^{tan θ} f (t sin θ) d t$
Put $t = x$ then we get,
$= - cos θ \int_{1}^{tan θ} f (x sin θ) d x$

Suggest Corrections

Similar questions

\int_{sin θ}^{cos θ} f (x tan θ) d x

(w h e r e θ \neq \frac{n π}{2}, n \in I)

Q. If

\cot θ - \tan θ = \sec θ

, then, θ is equal to
(a)

2 n π + \frac{3 π}{2}, n \in Z

(b)

n π + {(- 1)}^{n} \frac{π}{6}, n π Z

(c)

n π + \frac{π}{2}, n \in Z

(d) none of these.

Consider the system of linear equations in x, y and z ;
(sin 3 $θ$ ) x - y + z = 0 ......(i)
(cos 2 $θ$ )x + 4y + 3z = 0 ......(ii)
2x + 7y+ 7z = 0 ......(iii)

The value of $θ$ for which the system has nontrivial solution is

Which of the following statements are correct?

1 . if sin θ = sin α ⇒ θ = nπ + $(- 1)^{n}$ α where α ∈ [ - $\frac{π}{2}$ , $\frac{π}{2}$ ] n ∈ I

2 . if cos θ = cos α ⇒ 2nπ±α where α ∈ [0,π] n ∈ I

3 . if tan θ = tan α ⇒ θ = nπ + α where α ∈ (- $\frac{π}{2}$ , $\frac{π}{2}$ ) n ∈ I

Q. Solve

tan θ = 2

, then

θ = n π + a, w h e r e a = {tan}^{- 1} (2), n \in I

If true then enter

1

and if false then enter

0

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∫cosθsinθf(x tanθ) dx is (where θ≠nπ2,n∈I)

The correct option is A −cosθ∫tanθ1f(x sinθ)dxLet xcosθ=t⇒dx=(cosθ) dt ∫cosθsinθf(x tanθ) dx =∫1tanθf(t sinθ)cosθ dt=−cosθ∫tanθ1f(t sinθ)dt Put t=x then we get, =−cosθ∫tanθ1f(x sinθ)dx

$\int_{sin θ}^{cos θ} f (x tan θ) d x$ is
$(w h e r e θ \neq \frac{n π}{2}, n \in I)$