Math, asked by amanatsahibb, 1 month ago

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

Answers

Answered by PRINCE100001
12

Step-by-step explanation:

Question :

</p><p>\ \ \; \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

</p><p>\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</p><p>

\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} </p><p></p><p></p><p>

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

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

Answered by preetkumardujran85
3

Answer:

a minus x be answer.

Step-by-step explanation:

hope it helps you to

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