Integrate the function- -1-4 x-x2 d x

Let I=∫√(1 4x x^2)dx=∫√( (x^2+4x 1 2^2+2^2))dx =∫√( [(x+2)^2 (√(5))^2])dx =∫√((√(5))^2 (x+2)^2)dx [ ∵∫√(a^2 x^2)dx =x/2√(a^2 x^2)+a^2/2sin^ 1x/a+C ] ⇒ I= x+2/ ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

Integrate the...

Question

Integrate the function.
$\int \sqrt{1 - 4 x - x^{2}} d x .$

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Solution

Let $I = \int \sqrt{1 - 4 x - x^{2}} d x = \int \sqrt{- (x^{2} + 4 x - 1 - 2^{2} + 2^{2})} d x$
$= \int \sqrt{- [(x + 2)^{2} - (\sqrt{5})^{2}]} d x = \int \sqrt{(\sqrt{5})^{2} - (x + 2)^{2}} d x [∵ \int \sqrt{a^{2} - x^{2}} d x = \frac{x}{2} \sqrt{a^{2} - x^{2}} + \frac{a^{2}}{2} s i n^{- 1} \frac{x}{a} + C] \Rightarrow I = \frac{x + 2}{2} \sqrt{1 - 4 x - x^{2}} + \frac{5}{2} s i n^{- 1} \frac{(x + 2)}{\sqrt{5}} + C$

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Integrate the function. ∫√1−4x−x2dx.

Let I=∫√1−4x−x2dx=∫√−(x2+4x−1−22+22)dx =∫√−[(x+2)2−(√5)2]dx=∫√(√5)2−(x+2)2dx[∵∫√a2−x2dx=x2√a2−x2+a22sin−1xa+C]⇒I=x+22√1−4x−x2+52sin−1(x+2)√5+C

Integrate the function.
$\int \sqrt{1 - 4 x - x^{2}} d x .$