# Engineering Mathematics Questions and Answers – Integral Reduction Formula

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) π8
c) π4
d) $$\frac{15\pi}{96}$$

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) $$\frac{21\pi}{2048}$$

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

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) 110!
b) 5!6!11!
c) 10!5!6!
d) 0

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) 1513 + 11π4

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$$=\frac{1}{5}-\frac{1}{3}+\frac{1}{1}-\frac{\pi}{4}$$
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6. Find $$\int_{0}^{\frac{\pi}{4}} x^4.sin(x)dx$$.
a) -1
b) 1
c) 0
d) 4((π2)3 – 3π + 1)

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!

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) -16105

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.