Advertisements
Advertisements
प्रश्न
Evaluate the following:
`int_0^(pi/2) "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2` (Hint: Divide Numerator and Denominator by cos4x)
Advertisements
उत्तर
Let I = `int_0^(pi/2) "dx"/(("a"^2 cos^2x + "b"^2 sin^2 x)^2`
Dividing the numerator and denominator by cos4x, we have
I = `int_0^(pi/2) (sec^4x)/(("a"^2 cos^2x)/(cos^2x) + ("b"^2 sin^2x)/cos^2x)^2 "d"x`
= `int_0^(pi/2) (sec^2x * sec^2x)/("a"^2 + "b"^2 tan^2 x)^2 "d"x`
= `int_0^(pi/2) ((1 + tan^2x) sec^2x)/("a"^2 + "b"^2 tan^2 x)^2 "d"x`
Put tan x = t
⇒ sec2x dx = dt
Changing the limits, we get
When x = 0
t = tan 0 = 0
When x = `pi/2`
t = `tan pi/2 = oo`
∴ I = `int_0^oo (1 + "t"^2)/("a"^2 + "b"^2"t"^2)^2 "dt"`
Put t2 = u only for the purpose of partial fraction
∴ `(1 +"u")/("a"^2 + "b"^2"u")^2 = "A"/(("a"^2 + "b"^2"u")) + "B"/("a"^2 + "b"^2"u")^2`
1 + u = A(a2 + b2u) + B
Comparing the coefficients of like terms, we get
a2A + B = 1 and b2A = 1
⇒ A = `1/"b"^2`
Now `"a"^2 * 1/"b"^2 + "B"` = 1
⇒ B = `1 - "a"^2/"b"^2`
= `("b"^2 - "a"^2)/"b"^2`
∴ I = `int_0^oo (1 + "t"^2)/("a"^2 + "b"^2"t"^2)^2`
= `1/"b"^2 int_0^oo "dt"/("a"^2 + "b"^2"t"^2) + ("b"^2 - "a"^2)/"b"^2 int_0^oo "dt"/("a"^2 + "b"^2"t"^2)^2`
= `1/"b"^2 int_0^oo "dt"/("b"^2("a"^2/"b"^2 + "t"^2)) + ("b"^2 - "a"^2)/"b"^2 int_0^oo "dt"/("a"^2 + "b"^2"t"^2)^2`
= `1/"ab"^3 [tan^-1 "t"/("a"/"b")]_0^oo + ("b"^2 - "a"^2)/"b"^2 (pi/4 * 1/("a"^3"b"))`
= `1/"ab"^3 [tan^-1 oo - tan 0] + ("b"^2 - "a"^2)/"b"^2 (pi/(4"a"^3"b"))`
= `1/"ab"^3 * pi/2 + pi/4 * ("b"^2 - "a"^2)/("a"^2"b"^3)`
= `pi/(2"ab"^3) + pi/4 * ("b"^2 - "a"^2)/("a"^3"b"^3)`
= `pi [(2"a"^2 + "b"^2 - "a"^2)/(4"a"^3"b"^3)]`
= `pi/4 (("a"^2 + "b"^2)/("a"^3"b"^3))`
APPEARS IN
संबंधित प्रश्न
Prove that: `int_0^(2a)f(x)dx=int_0^af(x)dx+int_0^af(2a-x)dx`
`∫_4^9 1/sqrtxdx=`_____
(A) 1
(B) –2
(C) 2
(D) –1
Prove that `int_0^"a" "f" ("x") "dx" = int_0^"a" "f" ("a" - "x") "d x",` hence evaluate `int_0^pi ("x" sin "x")/(1 + cos^2 "x") "dx"`
Evaluate the following integrals : `int_2^5 sqrt(x)/(sqrt(x) + sqrt(7 - x))*dx`
`int_0^2 e^x dx` = ______.
`int_(-7)^7 x^3/(x^2 + 7) "d"x` = ______
By completing the following activity, Evaluate `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x`.
Solution: Let I = `int_2^5 (sqrt(x))/(sqrt(x) + sqrt(7 - x)) "d"x` ......(i)
Using the property, `int_"a"^"b" "f"(x) "d"x = int_"a"^"b" "f"("a" + "b" - x) "d"x`, we get
I = `int_2^5 ("( )")/(sqrt(7 - x) + "( )") "d"x` ......(ii)
Adding equations (i) and (ii), we get
2I = `int_2^5 (sqrt(x))/(sqrt(x) - sqrt(7 - x)) "d"x + ( ) "d"x`
2I = `int_2^5 (("( )" + "( )")/("( )" + "( )")) "d"x`
2I = `square`
∴ I = `square`
The c.d.f, F(x) associated with p.d.f. f(x) = 3(1- 2x2). If 0 < x < 1 is k`(x - (2x^3)/"k")`, then value of k is ______.
`int_-9^9 x^3/(4 - x^2)` dx = ______
`int_0^{pi/2} xsinx dx` = ______
`int_0^(pi/2) sqrt(cos theta) * sin^2 theta "d" theta` = ______.
f(x) = `{:{(x^3/k; 0 ≤ x ≤ 2), (0; "otherwise"):}` is a p.d.f. of X. The value of k is ______
`int_0^{1/sqrt2} (sin^-1x)/(1 - x^2)^{3/2} dx` = ______
`int_0^pi sin^2x.cos^2x dx` = ______
`int_(-1)^1 log ((2 - x)/(2 + x)) "dx" = ?`
The value of `int_2^7 (sqrtx)/(sqrt(9 - x) + sqrtx)dx` is ______
`int_0^pi x sin^2x dx` = ______
Evaluate `int_(-1)^2 "f"(x) "d"x`, where f(x) = |x + 1| + |x| + |x – 1|
`int_0^(2"a") "f"(x) "d"x = 2int_0^"a" "f"(x) "d"x`, if f(2a – x) = ______.
If `int (log "x")^2/"x" "dx" = (log "x")^"k"/"k" + "c"`, then the value of k is:
Evaluate: `int_0^(2π) (1)/(1 + e^(sin x)`dx
If `intxf(x)dx = (f(x))/2` then f(x) = ex.
If `int_0^K dx/(2 + 18x^2) = π/24`, then the value of K is ______.
If `int_0^1(3x^2 + 2x+a)dx = 0,` then a = ______
`int_1^2 x logx dx`= ______
`int_0^(2a)f(x)/(f(x)+f(2a-x)) dx` = ______
Evaluate the following integral:
`int_0^1x (1 - x)^5 dx`
Evaluate the following integral:
`int_-9^9 x^3/(4 - x^2) dx`
Evaluate the following integral:
`int_-9^9 x^3/(4-x^2)dx`
Solve the following.
`int_0^1e^(x^2)x^3dx`
Evaluate the following integral:
`int_0^1x(1 - x)^5dx`
\[\int_{-2}^{2}\left|x^{2}-x-2\right|\mathrm{d}x=\]
The value of \[\int_{-1}^{1}\left(\sqrt{1+x+x^{2}}-\sqrt{1-x+x^{2}}\right)\mathrm{d}x\] is
The area enclosed between the graph of y = x3 and the lines x = 0, y = 1, y = 8 is ______.
What is the value of a definite integral when its upper and lower limits are equal?
Select the formula for \[P_2\] : Splitting the Interval.
Under which condition does \[P_6\] give \[\int_{0}^{2a} f(x)\,dx=2\int_{0}^{a} f(x)\,dx\]?
