Question

❥Anushka࿐​​
Q1.integration of √ ( a² - x² ) dx is?​

Answer 1

Answer:

∫

a

2

−x

2

dx = ?

Solution :

\displaystyle \sf \int \sqrt{a^2-x^2}\ dx∫

a

2

−x

2

dx

Use Trigonometric sub. ,

Let , x = a sin θ

➠ dx = a cos θ dθ

\begin{gathered}\\ \longrightarrow \displaystyle \sf \int \sqrt{a^2-(a\ sin\ \theta)^2}\ (a\ cos\ \theta\ d \theta ) \\\end{gathered}

⟶∫

a

2

−(a sin θ)

2

(a cos θ dθ)

\longrightarrow \displaystyle \sf a\ \int \sqrt{a^2-a^2.sin^2 \theta}\ \ cos\ \theta\ d \theta⟶a ∫

a

2

−a

2

.sin

2

θ

cos θ dθ

\longrightarrow \displaystyle \sf a\ \int \sqrt{a^2(1-sin^2 \theta)}\ \ cos\ \theta\ d \theta⟶a ∫

a

2

(1−sin

2

θ)

cos θ dθ

\longrightarrow \displaystyle \sf a^2\ \int \sqrt{1-sin^2 \theta}\ \ cos\ \theta\ d \theta⟶a

2

∫

1−sin

2

θ

cos θ dθ

\longrightarrow \displaystyle \sf a^2\ \int \sqrt{cos^2 \theta}\ \ cos\ \theta\ d \theta⟶a

2

∫

cos

2

θ

cos θ dθ

\longrightarrow \displaystyle \sf a^2\ \int cos^2 \theta\ d \theta⟶a

2

∫cos

2

θ dθ

\longrightarrow \displaystyle \sf a^2\ \int \left( \dfrac{1+cos\ 2\theta}{2} \right)\ d \theta⟶a

2

∫(

2

1+cos 2θ

) dθ

\longrightarrow \displaystyle \sf \dfrac{a^2}{2}\ \left[ \theta + \dfrac{sin\ 2 \theta}{2}\right]+ c⟶

2

a

2

[θ+

2

sin 2θ

]+c

\longrightarrow \displaystyle \sf \dfrac{a^2}{2}\ \left[ \theta + sin\ \theta .cos\ \theta \right]+ c⟶

2

a

2

[θ+sin θ.cos θ]+c

\begin{gathered}\\ \longrightarrow \sf \dfrac{a^2}{2} \left[ sin^{-1}\ \dfrac{x}{a} + \dfrac{x}{a}.\dfrac{\sqrt{a^2-x^2}}{a} \right]+c \\\end{gathered}

⟶

2

a

2

[sin

−1

a

x

+

a

x

.

a

a

2

−x

2

]+c

\longrightarrow \sf \dfrac{a^2}{2} \left[ sin^{-1}\ \dfrac{x}{a} + \dfrac{x . \sqrt{a^2-x^2}}{a^2} \right]+c⟶

2

a

2

[sin

−1

a

x

+

a

2

x.

a

2

−x

2

]+c

\longrightarrow \sf \pink{\dfrac{a^2}{2}\ sin^{-1}\ \dfrac{x}{a} + \dfrac{x}{2}\ \sqrt{a^2-x^2} +c}⟶

2

a

2

sin

−1

a

x

+

2

x

a

2

−x

2

+c

★ ═════════════════════ ★

\displaystyle \sf \green{ \int \sqrt{a^2-x^2}\ dx = \dfrac{a^2}{2}\ sin^{-1}\ \dfrac{x}{a} + \dfrac{x}{2}\ \sqrt{a^2-x^2} +c}∫

a

2

−x

2

dx=

2

a

2

sin

−1

a

x

+

2

x

a

2

−x

2

+c

Step-by-step explanation:

this is your answer thank you for your question

Answer 2

Answer:

$\orange{ \sf\displaystyle \int \sqrt{ {a}^{2} - {x}^{2} } = \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + c}$

Step-by-step explanation:

$\bf \: given \\ \displaystyle \: \int \sqrt{ {a}^{2} - {x}^{2} } \: dx \\ \sf substitute \\ \sf\: \: x = asin \theta \: \implies \: \theta = {sin}^{ - 1} \frac{x}{a} \: \: \: and \\ \sf \: \pink{dx = acos\theta \: d\theta \: } \\ \bf \: now \\ \\ \displaystyle \: \int \sqrt{ {a}^{2} - {x}^{2} } \: dx \\ = \sf \int \sqrt{ {a}^{2} - {a}^{2} {sin}^{2} \theta} \: \: acos\theta \: d\theta \\ = \sf \int \sqrt{ {a}^{2} (1 - {sin}^{2} \theta)} \: \: acos\theta \: d\theta \\ = \sf \int \: a \sqrt{ {1 - {sin}^{2} \theta} } \: a cos\theta \: d\theta \\ = \sf \int {a}^{2} . {cos}^{2} \theta \: d\theta \: \\ = \sf {a}^{2} \int \: \frac{1}{2} (1 + cos2\theta) \: d\theta \: \: \: \: \sf \green{\{ \because \: 2 {cos}^{2} \theta = 1 + cos2\theta \}} \\ \sf= \frac{ {a}^{2} }{2} (\theta + \frac{sin2\theta}{2} ) + c \: \: \: \: \: \: \: \ \sf \red{ \{c = integral \: constant \}} \\ \\ \sf = \frac{ {a}^{2} }{2} (\theta + \frac{2sin\theta \: cos\theta}{2} ) + c \\ \\ \sf = \frac{ {a}^{2} }{2}(\theta + sin \theta \: \sqrt{1 - {sin}^{2}\theta } ) + c \\ \\ \sf = \frac{ {a}^{2} }{2} ( {sin}^{ - 1} \frac{x}{a} + \frac{x}{a} \sqrt{1 - \frac{ {x}^{2} }{ {a}^{2} } } ) + c \\ \\ \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{ {a}^{2} }{2} . \frac{x}{ {a}^{2} } \sqrt{ {a}^{2} - {x}^{2} } + c \\ \\ \sf = \frac{ {a}^{2} }{2} {sin}^{ - 1} \frac{x}{a} + \frac{x}{2} \sqrt{ {a}^{2} - {x}^{2} } + c$