Math, asked by amanatsahibb, 6 hours ago



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

❥Anushka࿐​​

Answers

Answered by raisanjeet8896
0

Answer will be:

x/2✓(a²-x²)+a²/2*sin -1(x/a)+c

Answered by PRINCE100001
13

Answer:

Step-by-step explanation:

Question :

\dagger\ \; \displaystyle \sf \red{\int \sqrt{a^2-x^2}\ dx\ =\ ?}† ∫

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

Similar questions