Q1.integration of √ ( a² - x² ) dx is?
Answers
Step-by-step explanation:
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:
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}∫
a
2
−x
2
==
2
a
2
sin
−1
a
x
+
2
x
a
2
−x
2
+c
Step-by-step explanation:
\begin{gathered} \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\end{gathered}
given
∫
a
2
−x
2
dx
substitute
x=asinθ⟹θ=sin
−1
a
x
and
dx=acosθdθ
now
∫
a
2
−x
2
dx
=∫
a
2
−a
2
sin
2
θ
acosθdθ
=∫
a
2
(1−sin
2
θ)
acosθdθ
=∫a
1−sin
2
θ
acosθdθ
=∫a
2
.cos
2
θdθ
=a
2
∫
2
1
(1+cos2θ)dθ{∵2cos
2
θ=1+cos2θ}
=
2
a
2
(θ+
2
sin2θ
)+c {c=integralconstant}
=
2
a
2
(θ+
2
2sinθcosθ
)+c
=
2
a
2
(θ+sinθ
1−sin
2
θ
)+c
=
2
a
2
(sin
−1
a
x
+
a
x
1−
a
2
x
2
)+c
=
2
a
2
sin
−1
a
x
+
2
a
2
.
a
2
x
a
2
−x
2
+c
=
2
a
2
sin
−1
a
x
+
2
x
a
2
−x
2
+c