Question

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

Answer 1

Step-by-step explanation:

Question :

$\dagger\ \; \displaystyle \sf \red{\int \sqrt{a^2-x^2}\ dx\ =\ ?}†</p><p>$

Solution :

$\displaystyle \sf \int \sqrt{a^2-x^2}\ dx</p><p>$

Use Trigonometric sub. ,

Let , x = a sin θ

➠ dx = a cos θ dθ

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

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

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

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

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

$</p><p>\longrightarrow \displaystyle \sf a^2\ \int cos^2 \theta\ d \theta$

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

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

$\longrightarrow \displaystyle \sf \dfrac{a^2}{2}\ \left[ \theta + sin\ \theta .cos\ \theta \right]+ c</p><p>$

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

$\longrightarrow \sf \dfrac{a^2}{2} \left[ sin^{-1}\ \dfrac{x}{a} + \dfrac{x . \sqrt{a^2-x^2}}{a^2} \right]+c</p><p> </p><p> </p><p> </p><p></p><p>$

$\longrightarrow \sf \pink{\dfrac{a^2}{2}\ sin^{-1}\ \dfrac{x}{a} + \dfrac{x}{2}\ \sqrt{a^2-x^2} +c} </p><p></p><p>$

★ ═════════════════════ ★

$\displaystyle \sf \green{ \int \sqrt{a^2-x^2}\ dx = \dfrac{a^2}{2}\ sin^{-1}\ \dfrac{x}{a} + \dfrac{x}{2}\ \sqrt{a^2-x^2} +c}</p><p> </p><p>$

Answer 2

Step-by-step explanation:

hello

yes of course

btw can i get ur intro