integration of √ ( a² - x² ) dx is?
Answers
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