86247
If \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=\operatorname{Atan}^{-1} \frac{x}{2}+B^{-1} \tan ^{-1}\left(\frac{x}{3}\right)+C,\) then \(\mathbf{A}-\mathbf{B}=\)
1 \(\frac{1}{6}\)
2 \(\frac{1}{30}\)
3 \(-\frac{1}{30}\)
4 \(-\frac{1}{6}\)
Explanation:
(A) : Given, \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=A \tan ^{-1} \frac{x}{2}+B \tan ^{-1}\left(\frac{x}{3}\right)+C\) L.H.S., \(I=\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x\) \(=\frac{1}{5} \int\left(\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right) d x\) \(=\frac{1}{5}\left[\frac{1}{2} \tan ^{-1} \frac{x}{2}-\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]+C\) \(=\frac{1}{10} \tan ^{-1} \frac{x}{2}-\frac{1}{15} \tan ^{-1} \frac{x}{3}+C\) On comparing L.H.S., we get - \(\mathrm{A}=\frac{1}{10}, \mathrm{~B}=\frac{-1}{15}\) \(\therefore \quad \mathrm{A}-\mathrm{B}=\frac{1}{10}-\left(-\frac{1}{15}\right)=\frac{5}{30}=\frac{1}{6}\)
MHT CET-2017
Integral Calculus
86248
If \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=\operatorname{acos} 8 x+C\), then \(a=\)
1 \(-\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 \(\frac{1}{16}\)
4 \(-\frac{1}{8}\)
Explanation:
(C) : Given, \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C\) L.H.S., \(I=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x\) \(=\int \frac{2 \cos ^{2} 4 x}{\sin ^{2} 2 x-\cos ^{2} 2 x} \cdot \sin 2 x \cdot \cos 2 x d x\) \(=-\int \sin 4 \mathrm{x} \cdot \cos 4 \mathrm{x} \mathrm{dx}\) \(=-\frac{1}{2} \int \sin 8 \mathrm{xdx}=\frac{1}{16} \cos 8 \mathrm{x}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get - \(\therefore \quad \mathrm{A}=\frac{1}{16}\)
(B) : Given, \(\int \frac{x^{3}-1}{x^{3}+x} d x=\int \frac{x^{3}+x-x-1}{x^{3}+x} d x\) \(=\int\left(1-\frac{x+1}{x\left(x^{2}+1\right)}\right) d x=\int d x-\int \frac{x+1}{x\left(x^{2}+1\right)} d x\) \(=\int d x-\int \frac{d x}{x^{2}+1}-\int \frac{d x}{x\left(x^{2}+1\right)}\) \(=x-\tan ^{-1} x-\int \frac{1}{x} d x-\int \frac{x}{x^{2}+1} d x\) \(=x-\tan ^{-1} x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)+C\) \(=x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)-\tan ^{-1} x+C\)
COMEDK-2015
Integral Calculus
86250
\(\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\)
1 \(\frac{2 \mathrm{e}-1}{\mathrm{e}}\)
2 \(\frac{e+1}{e}\)
3 \(\mathrm{e}-\frac{1}{\mathrm{e}}\)
4 \(\frac{1}{\mathrm{e}}\)
Explanation:
(C) : Let, \(I=\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\int_{-1}^{1} e^{x} \cdot d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\left[e^{x}\right]_{-1}^{1}\) \(\left[\because \int^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.\); if \(f(-x)=f(x)\) even function \(=0\), if \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\) odd function \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{27} \cdot \cos \mathrm{x}\) Let, \(\quad \mathrm{x}=-\mathrm{x}\) \(f(-x)=(-x)^{27} \cdot \cos (-x)\) \(=-x^{27} \cdot \cos x=-f(x) \text { odd function, }\) \(\therefore \quad \mathrm{I}=0+\left[\mathrm{e}^{\mathrm{x}}\right]_{-1}^{1}=\mathrm{e}^{1}-\mathrm{e}^{-1}=\mathrm{e}-\frac{1}{\mathrm{e}}\)
COMEDK-2016
Integral Calculus
86251
Evaluate \(\int_{1}^{3} \frac{\cos (\log x)}{x} d x\)
86247
If \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=\operatorname{Atan}^{-1} \frac{x}{2}+B^{-1} \tan ^{-1}\left(\frac{x}{3}\right)+C,\) then \(\mathbf{A}-\mathbf{B}=\)
1 \(\frac{1}{6}\)
2 \(\frac{1}{30}\)
3 \(-\frac{1}{30}\)
4 \(-\frac{1}{6}\)
Explanation:
(A) : Given, \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=A \tan ^{-1} \frac{x}{2}+B \tan ^{-1}\left(\frac{x}{3}\right)+C\) L.H.S., \(I=\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x\) \(=\frac{1}{5} \int\left(\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right) d x\) \(=\frac{1}{5}\left[\frac{1}{2} \tan ^{-1} \frac{x}{2}-\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]+C\) \(=\frac{1}{10} \tan ^{-1} \frac{x}{2}-\frac{1}{15} \tan ^{-1} \frac{x}{3}+C\) On comparing L.H.S., we get - \(\mathrm{A}=\frac{1}{10}, \mathrm{~B}=\frac{-1}{15}\) \(\therefore \quad \mathrm{A}-\mathrm{B}=\frac{1}{10}-\left(-\frac{1}{15}\right)=\frac{5}{30}=\frac{1}{6}\)
MHT CET-2017
Integral Calculus
86248
If \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=\operatorname{acos} 8 x+C\), then \(a=\)
1 \(-\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 \(\frac{1}{16}\)
4 \(-\frac{1}{8}\)
Explanation:
(C) : Given, \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C\) L.H.S., \(I=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x\) \(=\int \frac{2 \cos ^{2} 4 x}{\sin ^{2} 2 x-\cos ^{2} 2 x} \cdot \sin 2 x \cdot \cos 2 x d x\) \(=-\int \sin 4 \mathrm{x} \cdot \cos 4 \mathrm{x} \mathrm{dx}\) \(=-\frac{1}{2} \int \sin 8 \mathrm{xdx}=\frac{1}{16} \cos 8 \mathrm{x}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get - \(\therefore \quad \mathrm{A}=\frac{1}{16}\)
(B) : Given, \(\int \frac{x^{3}-1}{x^{3}+x} d x=\int \frac{x^{3}+x-x-1}{x^{3}+x} d x\) \(=\int\left(1-\frac{x+1}{x\left(x^{2}+1\right)}\right) d x=\int d x-\int \frac{x+1}{x\left(x^{2}+1\right)} d x\) \(=\int d x-\int \frac{d x}{x^{2}+1}-\int \frac{d x}{x\left(x^{2}+1\right)}\) \(=x-\tan ^{-1} x-\int \frac{1}{x} d x-\int \frac{x}{x^{2}+1} d x\) \(=x-\tan ^{-1} x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)+C\) \(=x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)-\tan ^{-1} x+C\)
COMEDK-2015
Integral Calculus
86250
\(\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\)
1 \(\frac{2 \mathrm{e}-1}{\mathrm{e}}\)
2 \(\frac{e+1}{e}\)
3 \(\mathrm{e}-\frac{1}{\mathrm{e}}\)
4 \(\frac{1}{\mathrm{e}}\)
Explanation:
(C) : Let, \(I=\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\int_{-1}^{1} e^{x} \cdot d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\left[e^{x}\right]_{-1}^{1}\) \(\left[\because \int^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.\); if \(f(-x)=f(x)\) even function \(=0\), if \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\) odd function \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{27} \cdot \cos \mathrm{x}\) Let, \(\quad \mathrm{x}=-\mathrm{x}\) \(f(-x)=(-x)^{27} \cdot \cos (-x)\) \(=-x^{27} \cdot \cos x=-f(x) \text { odd function, }\) \(\therefore \quad \mathrm{I}=0+\left[\mathrm{e}^{\mathrm{x}}\right]_{-1}^{1}=\mathrm{e}^{1}-\mathrm{e}^{-1}=\mathrm{e}-\frac{1}{\mathrm{e}}\)
COMEDK-2016
Integral Calculus
86251
Evaluate \(\int_{1}^{3} \frac{\cos (\log x)}{x} d x\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86247
If \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=\operatorname{Atan}^{-1} \frac{x}{2}+B^{-1} \tan ^{-1}\left(\frac{x}{3}\right)+C,\) then \(\mathbf{A}-\mathbf{B}=\)
1 \(\frac{1}{6}\)
2 \(\frac{1}{30}\)
3 \(-\frac{1}{30}\)
4 \(-\frac{1}{6}\)
Explanation:
(A) : Given, \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=A \tan ^{-1} \frac{x}{2}+B \tan ^{-1}\left(\frac{x}{3}\right)+C\) L.H.S., \(I=\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x\) \(=\frac{1}{5} \int\left(\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right) d x\) \(=\frac{1}{5}\left[\frac{1}{2} \tan ^{-1} \frac{x}{2}-\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]+C\) \(=\frac{1}{10} \tan ^{-1} \frac{x}{2}-\frac{1}{15} \tan ^{-1} \frac{x}{3}+C\) On comparing L.H.S., we get - \(\mathrm{A}=\frac{1}{10}, \mathrm{~B}=\frac{-1}{15}\) \(\therefore \quad \mathrm{A}-\mathrm{B}=\frac{1}{10}-\left(-\frac{1}{15}\right)=\frac{5}{30}=\frac{1}{6}\)
MHT CET-2017
Integral Calculus
86248
If \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=\operatorname{acos} 8 x+C\), then \(a=\)
1 \(-\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 \(\frac{1}{16}\)
4 \(-\frac{1}{8}\)
Explanation:
(C) : Given, \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C\) L.H.S., \(I=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x\) \(=\int \frac{2 \cos ^{2} 4 x}{\sin ^{2} 2 x-\cos ^{2} 2 x} \cdot \sin 2 x \cdot \cos 2 x d x\) \(=-\int \sin 4 \mathrm{x} \cdot \cos 4 \mathrm{x} \mathrm{dx}\) \(=-\frac{1}{2} \int \sin 8 \mathrm{xdx}=\frac{1}{16} \cos 8 \mathrm{x}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get - \(\therefore \quad \mathrm{A}=\frac{1}{16}\)
(B) : Given, \(\int \frac{x^{3}-1}{x^{3}+x} d x=\int \frac{x^{3}+x-x-1}{x^{3}+x} d x\) \(=\int\left(1-\frac{x+1}{x\left(x^{2}+1\right)}\right) d x=\int d x-\int \frac{x+1}{x\left(x^{2}+1\right)} d x\) \(=\int d x-\int \frac{d x}{x^{2}+1}-\int \frac{d x}{x\left(x^{2}+1\right)}\) \(=x-\tan ^{-1} x-\int \frac{1}{x} d x-\int \frac{x}{x^{2}+1} d x\) \(=x-\tan ^{-1} x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)+C\) \(=x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)-\tan ^{-1} x+C\)
COMEDK-2015
Integral Calculus
86250
\(\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\)
1 \(\frac{2 \mathrm{e}-1}{\mathrm{e}}\)
2 \(\frac{e+1}{e}\)
3 \(\mathrm{e}-\frac{1}{\mathrm{e}}\)
4 \(\frac{1}{\mathrm{e}}\)
Explanation:
(C) : Let, \(I=\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\int_{-1}^{1} e^{x} \cdot d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\left[e^{x}\right]_{-1}^{1}\) \(\left[\because \int^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.\); if \(f(-x)=f(x)\) even function \(=0\), if \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\) odd function \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{27} \cdot \cos \mathrm{x}\) Let, \(\quad \mathrm{x}=-\mathrm{x}\) \(f(-x)=(-x)^{27} \cdot \cos (-x)\) \(=-x^{27} \cdot \cos x=-f(x) \text { odd function, }\) \(\therefore \quad \mathrm{I}=0+\left[\mathrm{e}^{\mathrm{x}}\right]_{-1}^{1}=\mathrm{e}^{1}-\mathrm{e}^{-1}=\mathrm{e}-\frac{1}{\mathrm{e}}\)
COMEDK-2016
Integral Calculus
86251
Evaluate \(\int_{1}^{3} \frac{\cos (\log x)}{x} d x\)
86247
If \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=\operatorname{Atan}^{-1} \frac{x}{2}+B^{-1} \tan ^{-1}\left(\frac{x}{3}\right)+C,\) then \(\mathbf{A}-\mathbf{B}=\)
1 \(\frac{1}{6}\)
2 \(\frac{1}{30}\)
3 \(-\frac{1}{30}\)
4 \(-\frac{1}{6}\)
Explanation:
(A) : Given, \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=A \tan ^{-1} \frac{x}{2}+B \tan ^{-1}\left(\frac{x}{3}\right)+C\) L.H.S., \(I=\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x\) \(=\frac{1}{5} \int\left(\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right) d x\) \(=\frac{1}{5}\left[\frac{1}{2} \tan ^{-1} \frac{x}{2}-\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]+C\) \(=\frac{1}{10} \tan ^{-1} \frac{x}{2}-\frac{1}{15} \tan ^{-1} \frac{x}{3}+C\) On comparing L.H.S., we get - \(\mathrm{A}=\frac{1}{10}, \mathrm{~B}=\frac{-1}{15}\) \(\therefore \quad \mathrm{A}-\mathrm{B}=\frac{1}{10}-\left(-\frac{1}{15}\right)=\frac{5}{30}=\frac{1}{6}\)
MHT CET-2017
Integral Calculus
86248
If \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=\operatorname{acos} 8 x+C\), then \(a=\)
1 \(-\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 \(\frac{1}{16}\)
4 \(-\frac{1}{8}\)
Explanation:
(C) : Given, \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C\) L.H.S., \(I=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x\) \(=\int \frac{2 \cos ^{2} 4 x}{\sin ^{2} 2 x-\cos ^{2} 2 x} \cdot \sin 2 x \cdot \cos 2 x d x\) \(=-\int \sin 4 \mathrm{x} \cdot \cos 4 \mathrm{x} \mathrm{dx}\) \(=-\frac{1}{2} \int \sin 8 \mathrm{xdx}=\frac{1}{16} \cos 8 \mathrm{x}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get - \(\therefore \quad \mathrm{A}=\frac{1}{16}\)
(B) : Given, \(\int \frac{x^{3}-1}{x^{3}+x} d x=\int \frac{x^{3}+x-x-1}{x^{3}+x} d x\) \(=\int\left(1-\frac{x+1}{x\left(x^{2}+1\right)}\right) d x=\int d x-\int \frac{x+1}{x\left(x^{2}+1\right)} d x\) \(=\int d x-\int \frac{d x}{x^{2}+1}-\int \frac{d x}{x\left(x^{2}+1\right)}\) \(=x-\tan ^{-1} x-\int \frac{1}{x} d x-\int \frac{x}{x^{2}+1} d x\) \(=x-\tan ^{-1} x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)+C\) \(=x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)-\tan ^{-1} x+C\)
COMEDK-2015
Integral Calculus
86250
\(\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\)
1 \(\frac{2 \mathrm{e}-1}{\mathrm{e}}\)
2 \(\frac{e+1}{e}\)
3 \(\mathrm{e}-\frac{1}{\mathrm{e}}\)
4 \(\frac{1}{\mathrm{e}}\)
Explanation:
(C) : Let, \(I=\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\int_{-1}^{1} e^{x} \cdot d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\left[e^{x}\right]_{-1}^{1}\) \(\left[\because \int^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.\); if \(f(-x)=f(x)\) even function \(=0\), if \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\) odd function \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{27} \cdot \cos \mathrm{x}\) Let, \(\quad \mathrm{x}=-\mathrm{x}\) \(f(-x)=(-x)^{27} \cdot \cos (-x)\) \(=-x^{27} \cdot \cos x=-f(x) \text { odd function, }\) \(\therefore \quad \mathrm{I}=0+\left[\mathrm{e}^{\mathrm{x}}\right]_{-1}^{1}=\mathrm{e}^{1}-\mathrm{e}^{-1}=\mathrm{e}-\frac{1}{\mathrm{e}}\)
COMEDK-2016
Integral Calculus
86251
Evaluate \(\int_{1}^{3} \frac{\cos (\log x)}{x} d x\)
86247
If \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=\operatorname{Atan}^{-1} \frac{x}{2}+B^{-1} \tan ^{-1}\left(\frac{x}{3}\right)+C,\) then \(\mathbf{A}-\mathbf{B}=\)
1 \(\frac{1}{6}\)
2 \(\frac{1}{30}\)
3 \(-\frac{1}{30}\)
4 \(-\frac{1}{6}\)
Explanation:
(A) : Given, \(\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x=A \tan ^{-1} \frac{x}{2}+B \tan ^{-1}\left(\frac{x}{3}\right)+C\) L.H.S., \(I=\int \frac{1}{\left(x^{2}+4\right)\left(x^{2}+9\right)} d x\) \(=\frac{1}{5} \int\left(\frac{1}{x^{2}+4}-\frac{1}{x^{2}+9}\right) d x\) \(=\frac{1}{5}\left[\frac{1}{2} \tan ^{-1} \frac{x}{2}-\frac{1}{3} \tan ^{-1} \frac{x}{3}\right]+C\) \(=\frac{1}{10} \tan ^{-1} \frac{x}{2}-\frac{1}{15} \tan ^{-1} \frac{x}{3}+C\) On comparing L.H.S., we get - \(\mathrm{A}=\frac{1}{10}, \mathrm{~B}=\frac{-1}{15}\) \(\therefore \quad \mathrm{A}-\mathrm{B}=\frac{1}{10}-\left(-\frac{1}{15}\right)=\frac{5}{30}=\frac{1}{6}\)
MHT CET-2017
Integral Calculus
86248
If \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=\operatorname{acos} 8 x+C\), then \(a=\)
1 \(-\frac{1}{16}\)
2 \(\frac{1}{8}\)
3 \(\frac{1}{16}\)
4 \(-\frac{1}{8}\)
Explanation:
(C) : Given, \(\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x=a \cos 8 x+C\) L.H.S., \(I=\int \frac{\cos 8 x+1}{\tan 2 x-\cot 2 x} d x\) \(=\int \frac{2 \cos ^{2} 4 x}{\sin ^{2} 2 x-\cos ^{2} 2 x} \cdot \sin 2 x \cdot \cos 2 x d x\) \(=-\int \sin 4 \mathrm{x} \cdot \cos 4 \mathrm{x} \mathrm{dx}\) \(=-\frac{1}{2} \int \sin 8 \mathrm{xdx}=\frac{1}{16} \cos 8 \mathrm{x}+\mathrm{C}\) On comparing L.H.S. and R.H.S., we get - \(\therefore \quad \mathrm{A}=\frac{1}{16}\)
(B) : Given, \(\int \frac{x^{3}-1}{x^{3}+x} d x=\int \frac{x^{3}+x-x-1}{x^{3}+x} d x\) \(=\int\left(1-\frac{x+1}{x\left(x^{2}+1\right)}\right) d x=\int d x-\int \frac{x+1}{x\left(x^{2}+1\right)} d x\) \(=\int d x-\int \frac{d x}{x^{2}+1}-\int \frac{d x}{x\left(x^{2}+1\right)}\) \(=x-\tan ^{-1} x-\int \frac{1}{x} d x-\int \frac{x}{x^{2}+1} d x\) \(=x-\tan ^{-1} x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)+C\) \(=x-\log x+\frac{1}{2} \log \left(x^{2}+1\right)-\tan ^{-1} x+C\)
COMEDK-2015
Integral Calculus
86250
\(\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\)
1 \(\frac{2 \mathrm{e}-1}{\mathrm{e}}\)
2 \(\frac{e+1}{e}\)
3 \(\mathrm{e}-\frac{1}{\mathrm{e}}\)
4 \(\frac{1}{\mathrm{e}}\)
Explanation:
(C) : Let, \(I=\int_{-1}^{1}\left(x^{27} \cos x+e^{x}\right) d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\int_{-1}^{1} e^{x} \cdot d x\) \(=\int_{-1}^{1} x^{27} \cdot \cos x \cdot d x+\left[e^{x}\right]_{-1}^{1}\) \(\left[\because \int^{a} f(x) d x=2 \int_{0}^{a} f(x) d x\right.\); if \(f(-x)=f(x)\) even function \(=0\), if \(\mathrm{f}(-\mathrm{x})=-\mathrm{f}(\mathrm{x})\) odd function \(\therefore \quad \mathrm{f}(\mathrm{x})=\mathrm{x}^{27} \cdot \cos \mathrm{x}\) Let, \(\quad \mathrm{x}=-\mathrm{x}\) \(f(-x)=(-x)^{27} \cdot \cos (-x)\) \(=-x^{27} \cdot \cos x=-f(x) \text { odd function, }\) \(\therefore \quad \mathrm{I}=0+\left[\mathrm{e}^{\mathrm{x}}\right]_{-1}^{1}=\mathrm{e}^{1}-\mathrm{e}^{-1}=\mathrm{e}-\frac{1}{\mathrm{e}}\)
COMEDK-2016
Integral Calculus
86251
Evaluate \(\int_{1}^{3} \frac{\cos (\log x)}{x} d x\)