(A) : Given,\(I(m, n)=\int_{0}^{1} t^{m}(1+t)^{n} d t\) [We apply integration by parts, taking \((1+t)^{\mathrm{n}}\) as first and \(\mathrm{t}^{\mathrm{m}}\) as second function] \(I(m, n)=\left[(1+t)^{n} \cdot \frac{t^{m+1}}{m+1}\right]_{0}^{1}-\int_{0}^{1} n(1+t)^{n-1} \cdot \frac{t^{m+1}}{m+1} d t\) \(\quad=\frac{2^{n}}{m+1}-\frac{n}{m+1} \int_{0}^{1}(1+t)^{n-1} \cdot t^{m+1} d t\) \(\therefore I(m, n)=\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1)\)
Manipal-2012
Integral Calculus
86411
\(\int_{0}^{\pi / 4} \tan ^{2} x \sec ^{4} x d x=\)
1 \(1 / 3\)
2 \(4 / 15\)
3 1
4 \(8 / 15\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{4} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{2} \cdot \sec ^{2} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} d x\) Put, \(\tan x=t \Rightarrow \sec ^{2} x\). \(d x=d t\) When, \(x=0, t=0\) and \(x=\frac{\pi}{4}, t=1\) \(\therefore \mathrm{I}=\int_{0}^{1} \mathrm{t}^{2}\left(1+\mathrm{t}^{2}\right) \mathrm{dt}\) \(=\left[\frac{\mathrm{t}^{3}}{3}+\frac{\mathrm{t}^{5}}{5}\right]_{0}^{1}=\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}\)
(A) : Given,\(I(m, n)=\int_{0}^{1} t^{m}(1+t)^{n} d t\) [We apply integration by parts, taking \((1+t)^{\mathrm{n}}\) as first and \(\mathrm{t}^{\mathrm{m}}\) as second function] \(I(m, n)=\left[(1+t)^{n} \cdot \frac{t^{m+1}}{m+1}\right]_{0}^{1}-\int_{0}^{1} n(1+t)^{n-1} \cdot \frac{t^{m+1}}{m+1} d t\) \(\quad=\frac{2^{n}}{m+1}-\frac{n}{m+1} \int_{0}^{1}(1+t)^{n-1} \cdot t^{m+1} d t\) \(\therefore I(m, n)=\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1)\)
Manipal-2012
Integral Calculus
86411
\(\int_{0}^{\pi / 4} \tan ^{2} x \sec ^{4} x d x=\)
1 \(1 / 3\)
2 \(4 / 15\)
3 1
4 \(8 / 15\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{4} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{2} \cdot \sec ^{2} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} d x\) Put, \(\tan x=t \Rightarrow \sec ^{2} x\). \(d x=d t\) When, \(x=0, t=0\) and \(x=\frac{\pi}{4}, t=1\) \(\therefore \mathrm{I}=\int_{0}^{1} \mathrm{t}^{2}\left(1+\mathrm{t}^{2}\right) \mathrm{dt}\) \(=\left[\frac{\mathrm{t}^{3}}{3}+\frac{\mathrm{t}^{5}}{5}\right]_{0}^{1}=\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}\)
(A) : Given,\(I(m, n)=\int_{0}^{1} t^{m}(1+t)^{n} d t\) [We apply integration by parts, taking \((1+t)^{\mathrm{n}}\) as first and \(\mathrm{t}^{\mathrm{m}}\) as second function] \(I(m, n)=\left[(1+t)^{n} \cdot \frac{t^{m+1}}{m+1}\right]_{0}^{1}-\int_{0}^{1} n(1+t)^{n-1} \cdot \frac{t^{m+1}}{m+1} d t\) \(\quad=\frac{2^{n}}{m+1}-\frac{n}{m+1} \int_{0}^{1}(1+t)^{n-1} \cdot t^{m+1} d t\) \(\therefore I(m, n)=\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1)\)
Manipal-2012
Integral Calculus
86411
\(\int_{0}^{\pi / 4} \tan ^{2} x \sec ^{4} x d x=\)
1 \(1 / 3\)
2 \(4 / 15\)
3 1
4 \(8 / 15\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{4} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{2} \cdot \sec ^{2} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} d x\) Put, \(\tan x=t \Rightarrow \sec ^{2} x\). \(d x=d t\) When, \(x=0, t=0\) and \(x=\frac{\pi}{4}, t=1\) \(\therefore \mathrm{I}=\int_{0}^{1} \mathrm{t}^{2}\left(1+\mathrm{t}^{2}\right) \mathrm{dt}\) \(=\left[\frac{\mathrm{t}^{3}}{3}+\frac{\mathrm{t}^{5}}{5}\right]_{0}^{1}=\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}\)
(A) : Given,\(I(m, n)=\int_{0}^{1} t^{m}(1+t)^{n} d t\) [We apply integration by parts, taking \((1+t)^{\mathrm{n}}\) as first and \(\mathrm{t}^{\mathrm{m}}\) as second function] \(I(m, n)=\left[(1+t)^{n} \cdot \frac{t^{m+1}}{m+1}\right]_{0}^{1}-\int_{0}^{1} n(1+t)^{n-1} \cdot \frac{t^{m+1}}{m+1} d t\) \(\quad=\frac{2^{n}}{m+1}-\frac{n}{m+1} \int_{0}^{1}(1+t)^{n-1} \cdot t^{m+1} d t\) \(\therefore I(m, n)=\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1)\)
Manipal-2012
Integral Calculus
86411
\(\int_{0}^{\pi / 4} \tan ^{2} x \sec ^{4} x d x=\)
1 \(1 / 3\)
2 \(4 / 15\)
3 1
4 \(8 / 15\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{4} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{2} \cdot \sec ^{2} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} d x\) Put, \(\tan x=t \Rightarrow \sec ^{2} x\). \(d x=d t\) When, \(x=0, t=0\) and \(x=\frac{\pi}{4}, t=1\) \(\therefore \mathrm{I}=\int_{0}^{1} \mathrm{t}^{2}\left(1+\mathrm{t}^{2}\right) \mathrm{dt}\) \(=\left[\frac{\mathrm{t}^{3}}{3}+\frac{\mathrm{t}^{5}}{5}\right]_{0}^{1}=\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}\)
(A) : Given,\(I(m, n)=\int_{0}^{1} t^{m}(1+t)^{n} d t\) [We apply integration by parts, taking \((1+t)^{\mathrm{n}}\) as first and \(\mathrm{t}^{\mathrm{m}}\) as second function] \(I(m, n)=\left[(1+t)^{n} \cdot \frac{t^{m+1}}{m+1}\right]_{0}^{1}-\int_{0}^{1} n(1+t)^{n-1} \cdot \frac{t^{m+1}}{m+1} d t\) \(\quad=\frac{2^{n}}{m+1}-\frac{n}{m+1} \int_{0}^{1}(1+t)^{n-1} \cdot t^{m+1} d t\) \(\therefore I(m, n)=\frac{2^{n}}{m+1}-\frac{n}{m+1} \cdot I(m+1, n-1)\)
Manipal-2012
Integral Calculus
86411
\(\int_{0}^{\pi / 4} \tan ^{2} x \sec ^{4} x d x=\)
1 \(1 / 3\)
2 \(4 / 15\)
3 1
4 \(8 / 15\)
Explanation:
(D) : Let, \(I=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{4} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x \cdot \sec ^{2} \cdot \sec ^{2} x \cdot d x\) \(=\int_{0}^{\pi / 4} \tan ^{2} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} d x\) Put, \(\tan x=t \Rightarrow \sec ^{2} x\). \(d x=d t\) When, \(x=0, t=0\) and \(x=\frac{\pi}{4}, t=1\) \(\therefore \mathrm{I}=\int_{0}^{1} \mathrm{t}^{2}\left(1+\mathrm{t}^{2}\right) \mathrm{dt}\) \(=\left[\frac{\mathrm{t}^{3}}{3}+\frac{\mathrm{t}^{5}}{5}\right]_{0}^{1}=\frac{1}{3}+\frac{1}{5}=\frac{5+3}{15}=\frac{8}{15}\)