This set of Engineering Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Integral Reduction Formula”.
1. Find \(\int_{0}^{\frac{\pi}{2}} sin^6(x)dx\).
a) 0
b) π⁄8
c) π⁄4
d) \(\frac{15\pi}{96}\)
View Answer
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
=\(\frac{5.3.\pi}{6.4.2.2}\)
=\(\frac{15\pi}{96}\) .
2. Find \(\int_{0}^{\frac{\pi}{2}} sin^{10}(x)cos(x)dx\).
a) 1
b) 0
c) \(\frac{13\pi}{1098}\)
d) \(\frac{21\pi}{2048}\)
View Answer
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
\(\frac{9.7.5.3.\pi}{10.8.6.4.2.2}-\frac{11.9.7.5.3.\pi}{12.10.8.6.4.2.2}\)
\(\frac{9.7.5.3.\pi}{10.8.6.4.2.2}\times (1-\frac{11}{12})=\frac{21\pi}{2048}\)
3. Find \(\int_{0}^{\frac{\pi}{4}} tan^3(x)dx\).
a) 0
b) 1
c)-1
d) None of the mentioned
View Answer
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
View Answer
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
View Answer
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)
View Answer
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
\(\int_{0}^{\frac{\pi}{2}} x^4.sin(x)dx=4.\left ((\frac{\pi}{2})^{3}-3\int_0^{\frac{\pi}{2}} x^{2}sin(x)dx\right)\)
4((π/2)3-3π+1)
7. Find \(\int_{-\infty}^{0}x^5.e^x dx\).
a) 1
b) 199
c) -5!
d) 5!
View Answer
Explanation: Using the formula
\(\int_{-\infty}^{0}x^a.e^x \,dx=-a\int_{-\infty}^{0} x^{a-1}.e^x dx\)
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
View Answer
Explanation: Rewriting the function as
\(\int_{0}^{\frac{\pi}{2}} cos^3(x).cos(2x)dx=2\int_{0}^{\frac{\pi}{2}} cos^7(x)dx-\int_{0}^{\frac{\pi}{2}} cos^5(x)dx\)
We now use the formula
\(\int_0^{\frac{\pi}{2}} cos^n(x)dx=\frac{(n-1).(n-3)…2}{n.(n-2)…\times 1}\)
\(=2.\frac{6.4.2}{7.5.3.1}-\frac{4.2}{5.3.1}=\frac{4.2.-2}{5.3.1.7}\)
=\( \frac{16}{105}\)
Sanfoundry Global Education & Learning Series – Engineering Mathematics.
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