NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Integral Calculus
86148
If \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \left(\frac{x}{2}\right)+p \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) Then \(p\) is equal to
1 -4
2 4
3 2
4 -2
Explanation:
(A) : Given, \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \frac{x}{2}+p \cdot \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) \(=\int \frac{x-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x\) \(=-\frac{1}{2}\left[\int \frac{x}{\cos ^{2} \frac{x}{2}}-\int \frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x\right]\) \(=\left[\frac{1}{2} \int x \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(\frac{1}{2}\left[x \int \sec ^{2} \frac{x}{2} d x-\int \frac{d}{d x}(x) \int \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2}\left[x \frac{\tan \frac{x}{2}}{\frac{1}{2}}-\int \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2} \times 2 x \tan \frac{x}{2}-2 \times \frac{1}{2} \int \tan \frac{x}{2} d x-\int \tan \frac{x}{2} d x\) \(=\mathrm{x} \tan \frac{\mathrm{x}}{2}-2 \int \tan \frac{\mathrm{x}}{2} \mathrm{dx}\) \(=x \tan \frac{x}{2}-2\left[\log \frac{\left|\sec \frac{x}{2}\right|}{\frac{1}{2}}\right]+C\) \(=x \tan \frac{\mathrm{x}}{2}-4 \log \left|\sec \frac{\mathrm{x}}{2}\right|+\mathrm{C}\) So, \(\quad \mathrm{p}=-4\)
AP EAMCET-2013
Integral Calculus
86149
If \(\int \frac{\sin x}{\cos x(1+\cos x)} d x=f(x)+c\), then \(f(x)\) is equal to
1 \(\log \left|\frac{1+\cos x}{\cos x}\right|\)
2 \(\log \left|\frac{\cos x}{1+\cos x}\right|\)
3 \(\log \left|\frac{\sin x}{1+\sin x}\right|\)
4 \(\log \left|\frac{1+\sin x}{\sin x}\right|\)
Explanation:
(A) : Given, \(\int \frac{\sin x}{\cos x(1+\cos x)} d x\) Let, \(\cos \mathrm{x}=\mathrm{t}\) \(-\sin x d x=d t \Rightarrow I=-\int \frac{d t}{t(1+t)}\) \(\frac{-1}{t(1+t)}=\frac{A}{t}+\frac{B}{(1+t)}\) \(-1=\mathrm{A}+\mathrm{At}+\mathrm{Bt}\) \(-1=\mathrm{A}+(\mathrm{A}+\mathrm{B}) \mathrm{t}\) \(\mathrm{A}+\mathrm{B}=0\) and \(\mathrm{A}=-1\) \(\mathrm{A}=-\mathrm{B} \quad \mathrm{B}=1\) \(\int \frac{-1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}=-\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{\mathrm{t}+1} \mathrm{dt}\) \(=-[\log |t|-\log |t+1|]+c\) \(=-\log \left(\frac{\mathrm{t}}{\mathrm{t}+1}\right)+\mathrm{c}=\log \frac{\mathrm{t}+1}{\mathrm{t}}+\mathrm{c}\) \(=\log \left(\frac{\cos \mathrm{x}+1}{\cos \mathrm{x}}\right)+\mathrm{c}\) So, \(\quad f(x)=\log \left(\frac{\cos x+1}{\cos x}\right)\)
AP EAMCET-2005
Integral Calculus
86150
\(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\), then \(k\) is equal to
1 \(\frac{1}{50}\)
2 \(-\frac{1}{50}\)
3 \(\frac{1}{100}\)
4 \(-\frac{1}{100}\)
Explanation:
(C) : Given, \(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\) Let, \(I=\int \frac{x^{49} \cdot \tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^{2}} d x\) Put, \(\quad \mathrm{x}^{50}=\mathrm{t}\) \(50 x^{49} d x=d t \Rightarrow x^{49} d x=\frac{d t}{50}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \frac{\tan ^{-1} \mathrm{t}}{1+\mathrm{t}^{2}} \mathrm{dt}\) Again let \(\tan ^{-1} \mathrm{t}=\mathrm{v} \Rightarrow \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dv}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \mathrm{vdv}\) \(=\frac{1}{50} \times \frac{\mathrm{v}^{2}}{2}=\frac{1}{100}\left[\tan ^{-1} \mathrm{t}\right]^{2}=\frac{1}{100}\left[\tan ^{-1}\left(\mathrm{x}^{50}\right)\right]^{2}+\mathrm{c}\) So, \(\quad \mathrm{k}=\frac{1}{100}\)
86148
If \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \left(\frac{x}{2}\right)+p \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) Then \(p\) is equal to
1 -4
2 4
3 2
4 -2
Explanation:
(A) : Given, \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \frac{x}{2}+p \cdot \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) \(=\int \frac{x-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x\) \(=-\frac{1}{2}\left[\int \frac{x}{\cos ^{2} \frac{x}{2}}-\int \frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x\right]\) \(=\left[\frac{1}{2} \int x \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(\frac{1}{2}\left[x \int \sec ^{2} \frac{x}{2} d x-\int \frac{d}{d x}(x) \int \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2}\left[x \frac{\tan \frac{x}{2}}{\frac{1}{2}}-\int \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2} \times 2 x \tan \frac{x}{2}-2 \times \frac{1}{2} \int \tan \frac{x}{2} d x-\int \tan \frac{x}{2} d x\) \(=\mathrm{x} \tan \frac{\mathrm{x}}{2}-2 \int \tan \frac{\mathrm{x}}{2} \mathrm{dx}\) \(=x \tan \frac{x}{2}-2\left[\log \frac{\left|\sec \frac{x}{2}\right|}{\frac{1}{2}}\right]+C\) \(=x \tan \frac{\mathrm{x}}{2}-4 \log \left|\sec \frac{\mathrm{x}}{2}\right|+\mathrm{C}\) So, \(\quad \mathrm{p}=-4\)
AP EAMCET-2013
Integral Calculus
86149
If \(\int \frac{\sin x}{\cos x(1+\cos x)} d x=f(x)+c\), then \(f(x)\) is equal to
1 \(\log \left|\frac{1+\cos x}{\cos x}\right|\)
2 \(\log \left|\frac{\cos x}{1+\cos x}\right|\)
3 \(\log \left|\frac{\sin x}{1+\sin x}\right|\)
4 \(\log \left|\frac{1+\sin x}{\sin x}\right|\)
Explanation:
(A) : Given, \(\int \frac{\sin x}{\cos x(1+\cos x)} d x\) Let, \(\cos \mathrm{x}=\mathrm{t}\) \(-\sin x d x=d t \Rightarrow I=-\int \frac{d t}{t(1+t)}\) \(\frac{-1}{t(1+t)}=\frac{A}{t}+\frac{B}{(1+t)}\) \(-1=\mathrm{A}+\mathrm{At}+\mathrm{Bt}\) \(-1=\mathrm{A}+(\mathrm{A}+\mathrm{B}) \mathrm{t}\) \(\mathrm{A}+\mathrm{B}=0\) and \(\mathrm{A}=-1\) \(\mathrm{A}=-\mathrm{B} \quad \mathrm{B}=1\) \(\int \frac{-1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}=-\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{\mathrm{t}+1} \mathrm{dt}\) \(=-[\log |t|-\log |t+1|]+c\) \(=-\log \left(\frac{\mathrm{t}}{\mathrm{t}+1}\right)+\mathrm{c}=\log \frac{\mathrm{t}+1}{\mathrm{t}}+\mathrm{c}\) \(=\log \left(\frac{\cos \mathrm{x}+1}{\cos \mathrm{x}}\right)+\mathrm{c}\) So, \(\quad f(x)=\log \left(\frac{\cos x+1}{\cos x}\right)\)
AP EAMCET-2005
Integral Calculus
86150
\(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\), then \(k\) is equal to
1 \(\frac{1}{50}\)
2 \(-\frac{1}{50}\)
3 \(\frac{1}{100}\)
4 \(-\frac{1}{100}\)
Explanation:
(C) : Given, \(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\) Let, \(I=\int \frac{x^{49} \cdot \tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^{2}} d x\) Put, \(\quad \mathrm{x}^{50}=\mathrm{t}\) \(50 x^{49} d x=d t \Rightarrow x^{49} d x=\frac{d t}{50}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \frac{\tan ^{-1} \mathrm{t}}{1+\mathrm{t}^{2}} \mathrm{dt}\) Again let \(\tan ^{-1} \mathrm{t}=\mathrm{v} \Rightarrow \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dv}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \mathrm{vdv}\) \(=\frac{1}{50} \times \frac{\mathrm{v}^{2}}{2}=\frac{1}{100}\left[\tan ^{-1} \mathrm{t}\right]^{2}=\frac{1}{100}\left[\tan ^{-1}\left(\mathrm{x}^{50}\right)\right]^{2}+\mathrm{c}\) So, \(\quad \mathrm{k}=\frac{1}{100}\)
86148
If \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \left(\frac{x}{2}\right)+p \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) Then \(p\) is equal to
1 -4
2 4
3 2
4 -2
Explanation:
(A) : Given, \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \frac{x}{2}+p \cdot \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) \(=\int \frac{x-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x\) \(=-\frac{1}{2}\left[\int \frac{x}{\cos ^{2} \frac{x}{2}}-\int \frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x\right]\) \(=\left[\frac{1}{2} \int x \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(\frac{1}{2}\left[x \int \sec ^{2} \frac{x}{2} d x-\int \frac{d}{d x}(x) \int \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2}\left[x \frac{\tan \frac{x}{2}}{\frac{1}{2}}-\int \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2} \times 2 x \tan \frac{x}{2}-2 \times \frac{1}{2} \int \tan \frac{x}{2} d x-\int \tan \frac{x}{2} d x\) \(=\mathrm{x} \tan \frac{\mathrm{x}}{2}-2 \int \tan \frac{\mathrm{x}}{2} \mathrm{dx}\) \(=x \tan \frac{x}{2}-2\left[\log \frac{\left|\sec \frac{x}{2}\right|}{\frac{1}{2}}\right]+C\) \(=x \tan \frac{\mathrm{x}}{2}-4 \log \left|\sec \frac{\mathrm{x}}{2}\right|+\mathrm{C}\) So, \(\quad \mathrm{p}=-4\)
AP EAMCET-2013
Integral Calculus
86149
If \(\int \frac{\sin x}{\cos x(1+\cos x)} d x=f(x)+c\), then \(f(x)\) is equal to
1 \(\log \left|\frac{1+\cos x}{\cos x}\right|\)
2 \(\log \left|\frac{\cos x}{1+\cos x}\right|\)
3 \(\log \left|\frac{\sin x}{1+\sin x}\right|\)
4 \(\log \left|\frac{1+\sin x}{\sin x}\right|\)
Explanation:
(A) : Given, \(\int \frac{\sin x}{\cos x(1+\cos x)} d x\) Let, \(\cos \mathrm{x}=\mathrm{t}\) \(-\sin x d x=d t \Rightarrow I=-\int \frac{d t}{t(1+t)}\) \(\frac{-1}{t(1+t)}=\frac{A}{t}+\frac{B}{(1+t)}\) \(-1=\mathrm{A}+\mathrm{At}+\mathrm{Bt}\) \(-1=\mathrm{A}+(\mathrm{A}+\mathrm{B}) \mathrm{t}\) \(\mathrm{A}+\mathrm{B}=0\) and \(\mathrm{A}=-1\) \(\mathrm{A}=-\mathrm{B} \quad \mathrm{B}=1\) \(\int \frac{-1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}=-\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{\mathrm{t}+1} \mathrm{dt}\) \(=-[\log |t|-\log |t+1|]+c\) \(=-\log \left(\frac{\mathrm{t}}{\mathrm{t}+1}\right)+\mathrm{c}=\log \frac{\mathrm{t}+1}{\mathrm{t}}+\mathrm{c}\) \(=\log \left(\frac{\cos \mathrm{x}+1}{\cos \mathrm{x}}\right)+\mathrm{c}\) So, \(\quad f(x)=\log \left(\frac{\cos x+1}{\cos x}\right)\)
AP EAMCET-2005
Integral Calculus
86150
\(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\), then \(k\) is equal to
1 \(\frac{1}{50}\)
2 \(-\frac{1}{50}\)
3 \(\frac{1}{100}\)
4 \(-\frac{1}{100}\)
Explanation:
(C) : Given, \(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\) Let, \(I=\int \frac{x^{49} \cdot \tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^{2}} d x\) Put, \(\quad \mathrm{x}^{50}=\mathrm{t}\) \(50 x^{49} d x=d t \Rightarrow x^{49} d x=\frac{d t}{50}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \frac{\tan ^{-1} \mathrm{t}}{1+\mathrm{t}^{2}} \mathrm{dt}\) Again let \(\tan ^{-1} \mathrm{t}=\mathrm{v} \Rightarrow \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dv}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \mathrm{vdv}\) \(=\frac{1}{50} \times \frac{\mathrm{v}^{2}}{2}=\frac{1}{100}\left[\tan ^{-1} \mathrm{t}\right]^{2}=\frac{1}{100}\left[\tan ^{-1}\left(\mathrm{x}^{50}\right)\right]^{2}+\mathrm{c}\) So, \(\quad \mathrm{k}=\frac{1}{100}\)
86148
If \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \left(\frac{x}{2}\right)+p \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) Then \(p\) is equal to
1 -4
2 4
3 2
4 -2
Explanation:
(A) : Given, \(\int \frac{x-\sin x}{1+\cos x} d x=x \tan \frac{x}{2}+p \cdot \log \left|\sec \left(\frac{x}{2}\right)\right|+C\) \(=\int \frac{x-2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{2 \cos ^{2} \frac{x}{2}} d x\) \(=-\frac{1}{2}\left[\int \frac{x}{\cos ^{2} \frac{x}{2}}-\int \frac{2 \sin \frac{x}{2} \cdot \cos \frac{x}{2}}{\cos ^{2} \frac{x}{2}} d x\right]\) \(=\left[\frac{1}{2} \int x \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(\frac{1}{2}\left[x \int \sec ^{2} \frac{x}{2} d x-\int \frac{d}{d x}(x) \int \sec ^{2} \frac{x}{2} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2}\left[x \frac{\tan \frac{x}{2}}{\frac{1}{2}}-\int \frac{\tan \frac{x}{2}}{\frac{1}{2}} d x\right]-\int \tan \frac{x}{2} d x\) \(=\frac{1}{2} \times 2 x \tan \frac{x}{2}-2 \times \frac{1}{2} \int \tan \frac{x}{2} d x-\int \tan \frac{x}{2} d x\) \(=\mathrm{x} \tan \frac{\mathrm{x}}{2}-2 \int \tan \frac{\mathrm{x}}{2} \mathrm{dx}\) \(=x \tan \frac{x}{2}-2\left[\log \frac{\left|\sec \frac{x}{2}\right|}{\frac{1}{2}}\right]+C\) \(=x \tan \frac{\mathrm{x}}{2}-4 \log \left|\sec \frac{\mathrm{x}}{2}\right|+\mathrm{C}\) So, \(\quad \mathrm{p}=-4\)
AP EAMCET-2013
Integral Calculus
86149
If \(\int \frac{\sin x}{\cos x(1+\cos x)} d x=f(x)+c\), then \(f(x)\) is equal to
1 \(\log \left|\frac{1+\cos x}{\cos x}\right|\)
2 \(\log \left|\frac{\cos x}{1+\cos x}\right|\)
3 \(\log \left|\frac{\sin x}{1+\sin x}\right|\)
4 \(\log \left|\frac{1+\sin x}{\sin x}\right|\)
Explanation:
(A) : Given, \(\int \frac{\sin x}{\cos x(1+\cos x)} d x\) Let, \(\cos \mathrm{x}=\mathrm{t}\) \(-\sin x d x=d t \Rightarrow I=-\int \frac{d t}{t(1+t)}\) \(\frac{-1}{t(1+t)}=\frac{A}{t}+\frac{B}{(1+t)}\) \(-1=\mathrm{A}+\mathrm{At}+\mathrm{Bt}\) \(-1=\mathrm{A}+(\mathrm{A}+\mathrm{B}) \mathrm{t}\) \(\mathrm{A}+\mathrm{B}=0\) and \(\mathrm{A}=-1\) \(\mathrm{A}=-\mathrm{B} \quad \mathrm{B}=1\) \(\int \frac{-1}{\mathrm{t}(\mathrm{t}+1)} \mathrm{dt}=-\int \frac{1}{\mathrm{t}} \mathrm{dt}-\int \frac{1}{\mathrm{t}+1} \mathrm{dt}\) \(=-[\log |t|-\log |t+1|]+c\) \(=-\log \left(\frac{\mathrm{t}}{\mathrm{t}+1}\right)+\mathrm{c}=\log \frac{\mathrm{t}+1}{\mathrm{t}}+\mathrm{c}\) \(=\log \left(\frac{\cos \mathrm{x}+1}{\cos \mathrm{x}}\right)+\mathrm{c}\) So, \(\quad f(x)=\log \left(\frac{\cos x+1}{\cos x}\right)\)
AP EAMCET-2005
Integral Calculus
86150
\(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\), then \(k\) is equal to
1 \(\frac{1}{50}\)
2 \(-\frac{1}{50}\)
3 \(\frac{1}{100}\)
4 \(-\frac{1}{100}\)
Explanation:
(C) : Given, \(\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{\left(1+x^{100}\right)} d x=k\left(\tan ^{-1}\left(x^{50}\right)\right)^{2}+c\) Let, \(I=\int \frac{x^{49} \cdot \tan ^{-1}\left(x^{50}\right)}{1+\left(x^{50}\right)^{2}} d x\) Put, \(\quad \mathrm{x}^{50}=\mathrm{t}\) \(50 x^{49} d x=d t \Rightarrow x^{49} d x=\frac{d t}{50}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \frac{\tan ^{-1} \mathrm{t}}{1+\mathrm{t}^{2}} \mathrm{dt}\) Again let \(\tan ^{-1} \mathrm{t}=\mathrm{v} \Rightarrow \frac{1}{1+\mathrm{t}^{2}} \mathrm{dt}=\mathrm{dv}\) So, \(\quad \mathrm{I}=\frac{1}{50} \int \mathrm{vdv}\) \(=\frac{1}{50} \times \frac{\mathrm{v}^{2}}{2}=\frac{1}{100}\left[\tan ^{-1} \mathrm{t}\right]^{2}=\frac{1}{100}\left[\tan ^{-1}\left(\mathrm{x}^{50}\right)\right]^{2}+\mathrm{c}\) So, \(\quad \mathrm{k}=\frac{1}{100}\)
