Advertisements
Advertisements
Question
Find `int sqrt(10 - 4x + 4x^2) "d"x`
Advertisements
Solution
We have I = `int sqrt(10 - 4x + 4x^2) "d"x`
= `int sqrt((2x - 1)^2 + (3)^2) "d"x`
Put t = 2x – 1
Then dt = 2dx.
Therefore, I = `1/2 int sqrt("t"^2 + (3)^2) "dt"`
= `1/2 "t" sqrt("t"^2 + 9)/2 + 9/4 log|"t" + sqrt("t"^2 + 9)| + "C"`
= `1/4(2x - 1) sqrt((2x - 1)^2 + 9) + 9/4 log|(2x - 1) + sqrt((2x - 1)^2 + 9)| + "C"`
APPEARS IN
RELATED QUESTIONS
Evaluate the following integral:
Evaluate the following integral:
If \[f\left( x \right) = \int_0^x t\sin tdt\], the write the value of \[f'\left( x \right)\]
The value of \[\int\limits_0^{2\pi} \sqrt{1 + \sin\frac{x}{2}}dx\] is
Evaluate : \[\int\limits_0^{2\pi} \cos^5 x dx\] .
\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]
\[\int\limits_0^{\pi/4} \cos^4 x \sin^3 x dx\]
\[\int\limits_0^1 \log\left( 1 + x \right) dx\]
Evaluate the following integrals :-
\[\int_2^4 \frac{x^2 + x}{\sqrt{2x + 1}}dx\]
\[\int\limits_0^1 \left( \cos^{- 1} x \right)^2 dx\]
\[\int\limits_{- a}^a \frac{x e^{x^2}}{1 + x^2} dx\]
\[\int\limits_0^{\pi/2} \frac{1}{1 + \cot^7 x} dx\]
\[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} dx\]
\[\int\limits_0^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} dx\]
\[\int\limits_{- \pi}^\pi x^{10} \sin^7 x dx\]
\[\int\limits_{\pi/6}^{\pi/2} \frac{\ cosec x \cot x}{1 + {cosec}^2 x} dx\]
\[\int\limits_1^3 \left( 2 x^2 + 5x \right) dx\]
Prove that `int_a^b ƒ ("x") d"x" = int_a^bƒ(a + b - "x") d"x" and "hence evaluate" int_(π/6)^(π/3) (d"x")/(1+sqrt(tan "x")`
Evaluate the following:
Γ(4)
Choose the correct alternative:
The value of `int_(- pi/2)^(pi/2) cos x "d"x` is
Choose the correct alternative:
If n > 0, then Γ(n) is
Evaluate `int (3"a"x)/("b"^2 + "c"^2x^2) "d"x`
