Engineering Mathematics Multiple Choice Questions on “Integral Reduction Formula”.
1. Find (int_{0}^{frac{pi}{2}} sin^6(x)dx).
a) 0
b) π⁄8
c) π⁄4
d) (frac{15pi}{96})
Answer: d
Explanation: Using the formula for even n we have
(int_{0}^{frac{pi}{2}} sin^n(x)dx=frac{(n-1).(n-3)…3.pi}{n.(n-2)times …2.2})
We have
=(frac{5.3.pi}{6.4.2.2})
=(frac{15pi}{96})
.
2. Find (int_{0}^{frac{pi}{2}} sin^{10}(x)cos(x)dx).
a) 1
b) 0
c) (frac{13pi}{1098})
d) (frac{21pi}{2048})
Answer: d
Explanation: Rewriting the function as
=(int_{0}^{frac{pi}{2}} (sin^{10}(x)(1-sin^2(x)))dx=int_{0}^{frac{pi}{2}} sin^{10}(x)dx-int_{0}^{frac{pi}{2}} sin^{12}(x)dx)
We now apply the formula seperately for the two integrals
(frac{9.7.5.3.pi}{10.8.6.4.2.2}-frac{11.9.7.5.3.pi}{12.10.8.6.4.2.2})
(frac{9.7.5.3.pi}{10.8.6.4.2.2}times (1-frac{11}{12})=frac{21pi}{2048})
3. Find (int_{0}^{frac{pi}{4}} tan^3(x)dx).
a) 0
b) 1
c)-1
d) None of the mentioned
Answer: b
Explanation: Using the formula we have
(int_{0}^{frac{pi}{4}} tan^n(x)dx=frac{1}{n-1}-int_{0}^{frac{pi}{2}} tan^{n-2}(x)dx)
=(frac{1}{2}-int_{0}^{frac{pi}{4}} tan(x)dx)
=(frac{1-ln(2)}{2})
4. Find the value of (int_{0}^{frac{pi}{2}} cos^{11}(x).sin^9(x), dx).
a) 1⁄10!
b) 5!6!⁄11!
c) 10!⁄5!6!
d) 0
Answer: b
Explanation: Using the definition of beta function we see that the integral is equal to the beta function at (6,5)
Now using the relation between the Beta and the Gamma function we have
(beta(m, n)=frac{Gamma (m).Gamma (n)}{Gamma (m+n)})
(beta(6, 5)=frac{Gamma (6).Gamma (5)}{Gamma (11)}=frac{6!.5!}{11!})
5. Find (int_1^{sqrt{2}} (x^{frac{8}{5}}-frac{1}{x^{frac{2}{5}}})^{frac{5}{2}}dx).
a) -1
b) 1
c) 0
d) 1⁄5 – 1⁄3 + 1⁄1 – π⁄4
Answer: d
Explanation: Simplifying we have
(int_1^{sqrt{2}} (frac{x^2-1}{x})^{frac{5}{2}}dx)
Substitute x=sec(t)
(int_{0}^{frac{pi}{4}} tan^6(t)dt)
Now using the formula
(int_{0}^{frac{pi}{4}} tan^n(x)dx=frac{1}{n-1}-int_{0}^{frac{pi}{2}} tan^{n-2}(x)dx)
We have(=frac{1}{5}-frac{1}{3}+frac{1}{1}-frac{pi}{4})
6. Find (int_{0}^{frac{pi}{4}} x^4.sin(x)dx).
a) -1
b) 1
c) 0
d) 4((π⁄2)3 – 3π + 1)
Answer: b
Explanation: Using the formula
(int_{0}^{frac{pi}{2}} x^a.sin(x)dx=left ((frac{pi}{2})^{a-1}-(a-1)int x^{a-2}sin(x)right))
We have
(int_{0}^{frac{pi}{2}} x^4.sin(x)dx=4.left ((frac{pi}{2})^{3}-3int_0^{frac{pi}{2}} x^{2}sin(x)dxright))
4((π/2)3-3π+1)
7. Find (int_{-infty}^{0}x^5.e^x dx).
a) 1
b) 199
c) -5!
d) 5!
Answer: c
Explanation: Using the formula
(int_{-infty}^{0}x^a.e^x ,dx=-aint_{-infty}^{0} x^{a-1}.e^x dx)
We have
(int_{-infty}^{0}x^5.e^x =-5.-4.-3.-2.-1=-5!)
8. Find (int_{0}^{frac{pi}{2}} cos^3(x).cos(2x)dx).
a) 0
b) 5
c) 87
d) -16⁄105
Answer: d
Explanation: Rewriting the function as
(int_{0}^{frac{pi}{2}} cos^3(x).cos(2x)dx=2int_{0}^{frac{pi}{2}} cos^7(x)dx-int_{0}^{frac{pi}{2}} cos^5(x)dx)
We now use the formula
(int_0^{frac{pi}{2}} cos^n(x)dx=frac{(n-1).(n-3)…2}{n.(n-2)…times 1})
(=2.frac{6.4.2}{7.5.3.1}-frac{4.2}{5.3.1}=frac{4.2.-2}{5.3.1.7})
=( frac{16}{105})