-sin-cos- f x tan- d x is- where -n -2- n - IA- -cos-1tan- f x sin- dxB- -tan-sin-cos- f x dxC- sin-0tan- f x cos- dxD- -sin-sin-tan- f x dx

The correct option is A cosθ∫_1^tanθ f(x sinθ)dxLet x/cosθ = t ⇒ dx = (cosθ) dt ∫_sinθ^cosθ f(x tanθ) dx =∫^1_tanθ f(t sinθ) cosθ dt = cosθ∫_1^tanθ f(t si ...

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Byju's Answer

Standard XII

Mathematics

Integration by Substitution

∫sinθcosθ f x...

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

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Similar questions

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. Prove that :

\frac{tan θ}{1 - tan θ} - \frac{cot θ}{1 - cot θ} = \frac{cos θ + sin θ}{cos θ - sin θ}

Q. Find if

\frac{sin θ - cos θ + 1}{sin θ + cos θ + 1} = \frac{1}{sec θ - tan θ}

using the identity

{sec}^{2} θ = 1 + {tan}^{2} θ

\frac{cos θ}{sin θ} \times \frac{tan θ}{csc θ}

\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)

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