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BCom programme at NIU's School of Business Management is more affordable with a total tuition fee of INR 2.5 Lacs and a strong Shiksha Rating of 4.8/5. In comparison, Sharda and Galgotias offer higher fees and slightly lower ratings, making NIU a value-for-money option with strong academic credentials.

IBS Bangalore offers good placements for its MBA course annually. During IBS Bangalore placement 2025, 90% of students of the batch 2025 were placed in reputed companies. The key highlights of IBS Bangalore placements for the Class of 2024 and 2025 are tabulated below:

This is a Short answer type Questions as classified in NCERT Exemplar

Vidyavardhini College of Engineering and Technology offers courses such as BTech for 4 years. Admission to the respective courses is offered on the basis of minimum aggregate of 45% (40% in case of SC/ST/OBC candidates from Maharashtra) in the qualifying exam. Final admission is offered based on the candidates' MHT-CET/ JEE Main scores. The BTech fee at Vidyavardhini's College of Engineering and Technology includes tuition fee, development fee, admission process fee, etc. The tuition fee for BTech at VCET Mumbai is INR 4.34 Lacs for 4 years.

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{si{n}^{6}x+co{s}^{6}x}{si{n}^{2}x.co{s}^{2}x}dx}\\ I={\displaystyle \int \frac{{\left(si{n}^{2}x+co{s}^{2}x\right)}^3-3si{n}^{2}xco{s}^{2}x\left(si{n}^{2}x+co{s}^{2}x\right)}{si{n}^{2}x.co{s}^{2}x}dx}\\ \left[?a^3+b^3={\left(a+b\right)}^3-3ab\left(a+b\right)\right]\\ ={\displaystyle \int \frac{{\left(1\right)}^3-3si{n}^{2}xco{s}^{2}x.\left(1\right)}{si{n}^{2}x.co{s}^{2}x}}dx\\ ={\displaystyle \int \frac{1-3si{n}^{2}xco{s}^{2}x}{si{n}^{2}x.co{s}^{2}x}}dx\\ ={\displaystyle \int \left(\frac{1}{si{n}^{2}x.co{s}^{2}x}-\frac{3si{n}^{2}xco{s}^{2}x}{si{n}^{2}x.co{s}^{2}x}\right)}dx\\ ={\displaystyle \int \left(\frac{1}{si{n}^{2}x.co{s}^{2}x}-3\right)}dx={\displaystyle \int \left(\frac{si{n}^{2}x+co{s}^{2}x}{si{n}^{2}x.co{s}^{2}x}-3\right)}dx\\ ={\displaystyle \int \left[\left(\frac{1}{co{s}^{2}x}+\frac{1}{si{n}^{2}x}\right)-3\right]}dx\\ ={\displaystyle \int \left(se{c}^{2}x+cose{c}^{2}x-3\right)dx}\\ ={\displaystyle \int se{c}^{2}xdx+{\displaystyle \int cose{c}^{2}xdx-3{\displaystyle \int 1dx}}}\\ =tanx-cotx-3x+C\\ Hence,I=tanx-cotx-3x+C.\end{array}$

The college offers BTech, MTech, and MBA courses at the UG and PG levels. Apart from the UG and PG courses, the college also offers PhD courses. The selection criteria are based on entrance exam scores. Candidates looking for course admission at CEC can check the table below to know detailed selection criteria:

Candidates belonging to Scheduled Caste and Scheduled Tribe who have given consent for availing free travel facility only can avail 'Free Travel Authority' (free Railway Pass).

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{cos5x+cos4x}{1-2cos3x}dx}\\ I={\displaystyle \int \frac{2cos\frac{5x+4x}{2}.cos\frac{5x-4x}{2}}{1-2\left(co{s}^2\frac{3x}{2}-1\right)}dx}\\ ={\displaystyle \int \frac{2cos\frac{9x}{2}.cos\frac{x}{2}}{1-4co{s}^2\frac{3x}{2}+2}}dx={\displaystyle \int \frac{2cos\frac{9x}{2}.cos\frac{x}{2}}{3-4co{s}^2\frac{3x}{2}}dx}\\ =-{\displaystyle \int \frac{2cos\frac{9x}{2}.cos\frac{x}{2}}{4co{s}^2\frac{3x}{2}-3}dx}=-{\displaystyle \int \frac{2cos\frac{9x}{2}.cos\frac{x}{2}.cos\frac{3x}{2}}{4co{s}^3\frac{3x}{2}-3cos\frac{3x}{2}}dx}\left[\begin{array}{l}Multiplyinganddividing\\ bycos\frac{3x}{2}\end{array}\right]\\ =-{\displaystyle \int \frac{2cos\frac{9x}{2}.cos\frac{x}{2}.cos\frac{3x}{2}}{cos3.\frac{3x}{2}}dx}\left[?cos3x=4co{s}^{3}x-3cosx\right]\\ =-{\displaystyle \int \frac{2cos\frac{9x}{2}.cos\frac{x}{2}.cos\frac{3x}{2}}{cos\frac{9x}{2}}dx}=-{\displaystyle \int 2cos\frac{3x}{2}.cos\frac{x}{2}dx}\\ =-{\displaystyle \int \left[cos\left(\frac{3x}{2}+\frac{x}{2}\right)+cos\left(\frac{3x}{2}-\frac{x}{2}\right)\right]}dx\\ =-{\displaystyle \int \left(cos2x+cosx\right)dx\left[?2cosAcosB=cos\left(A+B\right)+cos\left(A-B\right)\right]}\\ =-{\displaystyle \int cos2xdx-{\displaystyle \int cosxdx}}=-\frac{1}{2}sin2x-sinx+C\\ Hence,I=-\left[\frac{1}{2}sin2x+sinx\right]+C\end{array}$

Some popular scholarships for international students with their eligibility are:

This is a Short answer type Questions as classified in NCERT Exemplar

This is a Short answer type Questions as classified in NCERT Exemplar

The primary eligibility criteria required for admission to BTech, MTech, and MBA courses at Chouksey Engineering College must meet the eligibility criteria set by the college. Students can check the table below to know the detailed course-wise eligibility:

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x^2}{1-x^4}dx}\\ ={\displaystyle \int \frac{x^2}{\left(1-x^2\right)\left(1+x^2\right)}dx}\\ Put{x}^2=tforthepurposeofpartialfractions.\\ Weget\frac{t}{\left(1-t\right)\left(1+t\right)}\\ \text{\hspace{0.17em}Resolvinginto}partialfractionsweput\\ \frac{t}{\left(1-t\right)\left(1+t\right)}=\frac{A}{1-t}+\frac{B}{1+t}\left[WhereAandBarearbitary\right]\\ \Rightarrow \frac{t}{\left(1-t\right)\left(1+t\right)}=\frac{A\left(1+t\right)+B\left(1-t\right)}{\left(1-t\right)\left(1+t\right)}\\ \Rightarrow t=A+At+B-Bt\\ Comparingtheliketerms,wegetA-B=1andA+B=0\\ Solvingtheaboveequations,wehaveA=\frac{1}{2}andB=-\frac{1}{2}\\ \therefore I={\displaystyle \int \frac{\frac{1}{2}}{1-x^2}}dx+{\displaystyle \int \frac{-\frac{1}{2}}{1+x^2}}dx\left[Puttingt=x^2\right]\\ =\frac{1}{2}.\frac{1}{2.1}log\left|\frac{1+x}{1-x}\right|-\frac{1}{2}ta{n}^{-1}x+C\\ =\frac{1}{4}log\left|\frac{1+x}{1-x}\right|-\frac{1}{2}ta{n}^{-1}x+C\\ Hence,I=\frac{1}{4}log\left|\frac{1+x}{1-x}\right|-\frac{1}{2}ta{n}^{-1}x+C.\end{array}$

This is a Short answer type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetI={\displaystyle \int \frac{x}{x^4-1}dx}\\ Put{x}^2=t\Rightarrow 2xdx=dt\Rightarrow xdx=\frac{dt}{2}\\ =\frac{1}{2}{\displaystyle \int \frac{dt}{t^2-1}}=\frac{1}{2}{\displaystyle \int \frac{dt}{t^2-{\left(1\right)}^2}}=\frac{1}{2}.\frac{1}{2.1}log\left|\frac{t-1}{t+1}\right|+C\\ \left[?\frac{1}{2}{\displaystyle \int \frac{dt}{x^2-a^2}}=\frac{1}{2a}log\left|\frac{x-a}{x+a}\right|+C\right]\\ =\frac{1}{4}log\left|\frac{x^2-1}{x^2+1}\right|+C\\ Hence,I=\frac{1}{4}log\left|\frac{x^2-1}{x^2+1}\right|+C.\end{array}$

This is a Short answer type Questions as classified in NCERT Exemplar