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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.10) Exercise 7.10

The Central Board for Secondary Education CBSE is one of the most highly esteemed and biggest educational boards in India, with many schools affiliated with it. CBSE is a central board, and its schools are present all over the country. The educational and academic approach implemented is one that has a very objective style. Therefore, students are required to have a very deep understanding of the subject, which is highly essential for them to score well in the subject during the exams. The CBSE board is widely preferred by students who are trying to prepare for and excel in competitive exams, as these require the same in-depth knowledge and level of preparation as the Class 12 board exams. Since the objective pattern of the CBSE matches completely with the objective question pattern in the competitive exams, the CBSE is suitable for the students who are appearing for such competitive exams.

The National Council of Educational Research and Training (NCERT) is a resource-based organisation that legislates every issue that is related to academic development and educational matters in the country. It works alongside the central government as well as the state governments. The NCERT and CBSE work together to formulate the curriculum that CBSE prescribes for all its students. CBSE strictly follows all the NCERT guidelines, their syllabus, and their prescribed exercises. The books that are prescribed by NCERT are used by the students over their entire academic course, and students are advised to use the NCERT textbooks as their main book of reference.

The CBSE curriculum is known to have a reputation for having one of the most massive syllabuses. It is very difficult for a student to try to complete the syllabus and adjust to the new academic environment. Practising with the help of the NCERT Solutions Class 12 Maths Chapter 7 Exercise 7.10, allows students to familiarise themselves with the paper pattern. After consulting with teachers, it can be established that one of the ways students can improve is through consistent practise. Students are advised to inculcate a habit of regular practice and revision in their preparation schedule.

Extramarks allows students to access the NCERT Solutions Class 12 Maths Chapter 7 Exercise 7.10, which solve all the problems in the concerned chapter discussed above. With the solutions, students seem to highly benefit from getting acquainted with the most significant topics. The explanations and methods in the answers are easy to understand and easily available on the Extramarks website. Practising the NCERT Solutions Class 12 Maths Chapter 7 Exercise 7.10 gives students an in-depth knowledge of the core concepts of the chapter for students to score well during their board exams.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.10) Exercise 7.10

NCERT Solutions aims to make the learning process for students as convenient as possible. Therefore, Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, which give the solutions to every unsolved question in the NCERT textbooks. The concepts discussed in Class 12 NCERT Solutions Maths Chapter 7 Exercise 7.10 can be somewhat challenging to understand, so practising is a prerequisite.

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, are made available through the combined efforts of qualified educators in the field of mathematics. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, are compiled and curated by Extramarks experts. Often, it has been observed that the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 provide different approaches to solving a problem, and in these, the provided solution has been chosen to be the most convenient and easy way for most of the students to follow because this way it can be easily understood by as many students as possible. After compilation and before going live on the website, the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 are reviewed multiple times, and then they are finally made accessible to students. This extensive process is followed to ensure that all the solutions provided in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 are correct.

Given the advanced level of difficulty, it is understandable that there will be many students with doubts regarding several steps that are involved in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10. To overcome this Extramarks provides lucid and easy explanations for all the steps involved in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10.

Access NCERT Solutions for Class 12 Maths Chapter 7 – Integrals

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, are released as a solution key to the Chapter 12 CBSE Integrals. The link to the PDF is available online on the Extramarks website, and with it, there are other links available to other exercises, chapters, and subjects across all CBSE classes.

The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 can be accessed by any student who accesses the Extramarks website searching for learning resources. All necessary information is provided on the Extramarks website and also on the mobile application.

What does the NCERT Class 12 Chapter 7 Exercise 7.10 Contain?

As can be observed in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, Chapter 12 is based on integrals, and Exercise 7.10 is based completely on definite integrals but being the last exercise of the chapter, it demands the implementation of concepts learned in the chapter from the beginning. When a student goes through the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, they can observe that the solutions in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10 are achieved through other concepts like limits, differentiation, trigonometric identities, and inverse trigonometric identities.

Chapter 12 is one of the most important chapters in the entire syllabus, and it implements and incorporates a plethora of formulas into the wide knowledge base of Integrals that exists for students. The formulas learned in this chapter and other chapters are tactfully provided in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10. Anyone can find all formulas written in one place for the student to revise the formulas easily in one go.

Q.1

$Evaluate the integrals \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} \cos^{5} ϕ dϕ$

Ans

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} \cos^{5} ϕ d ϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} \cos^{4} ϕ cosϕ dϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} {(\cos^{2} ϕ)}^{2} cosϕ dϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} {(1 - \sin^{2} ϕ)}^{2} cosϕ dϕ \\ = \int_{0}^{\frac{π}{2}} \sqrt{sinϕ} (1 - 2 \sin^{2} ϕ + \sin^{4} ϕ) cosϕ dϕ \\ Let t = sinϕ \Rightarrow \frac{dt}{dϕ} = cosϕ \\ When ϕ = 0, t = 0 and when ϕ = \frac{π}{2}, t = 1 \\ ∴ I = \int_{0}^{1} \sqrt{t} (1 - 2 t^{2} + t^{4}) dt \\ = \int_{0}^{1} (t^{\frac{1}{2}} - 2 t^{\frac{5}{2}} + t^{\frac{9}{2}}) dt \\ = {[\frac{t^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{t^{\frac{7}{2}}}{\frac{7}{2}} + \frac{t^{\frac{11}{2}}}{\frac{11}{2}}]}_{0}^{1} \\ = [\frac{1^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{1^{\frac{7}{2}}}{\frac{7}{2}} + \frac{1^{\frac{11}{2}}}{\frac{11}{2}}] - [\frac{0^{\frac{3}{2}}}{\frac{3}{2}} - 2 \frac{0^{\frac{7}{2}}}{\frac{7}{2}} + \frac{0^{\frac{11}{2}}}{\frac{11}{2}}] \\ = [\frac{2}{3} - 2 . \frac{2}{7} + \frac{2}{11}] \\ = \frac{2}{3} - \frac{4}{7} + \frac{2}{11} \\ = \frac{154 - 132 + 42}{231} = \frac{64}{231} \end{array}$

Q.3

$Evaluate the integrals \int_{0}^{1} \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx$

Ans

$\begin{array}{l} Let I = \int_{0}^{1} \sin^{- 1} (\frac{2 x}{1 + x^{2}}) dx \\ Also, let x = tan θ \Rightarrow \frac{dx}{dθ} = {sec}^{2} θ \\ When x = 0, θ = 0 and when x = 1, θ = \frac{π}{4} \\ ∴ I = \int_{0}^{\frac{π}{4}} {sin}^{- 1} (\frac{2 tan θ}{1 + {tan}^{2} θ}) {sec}^{2} θ d θ \\ = \int_{0}^{\frac{π}{4}} {sin}^{- 1} (sin 2 θ) {sec}^{2} θdθ \\ = \int_{0}^{\frac{π}{4}} 2 θ {sec}^{2} θ dθ \\ Now, \\ \int 2 θ {sec}^{2} θ dθ \\ = 2 [θ \int {sec}^{2} θ dθ - \int (\frac{d}{dθ} θ \int \sec^{2} θ dθ) dθ] \\ = 2 [θ tan θ - \int (1 . tan θ) dθ] \\ = 2 [θ tan θ + log cos θ] \\ ∴ I = 2 {[θ tan θ + log cos θ]}_{0}^{\frac{π}{4}} \\ = 2 [(\frac{π}{4}) tan \frac{π}{4} + log cos \frac{π}{4}] - [2 (0) tan 0 + log cos 0] \\ = 2 [\frac{π}{4} \times 1 + log \frac{1}{\sqrt{2}}] - [0 + log 1] \\ = [\frac{π}{2} - 2 \times \frac{1}{2} log 2] - [0 + 0] \\ = (\frac{π}{2} - log 2) \end{array}$

Q.4

$Evaluate the integrals \int_{0}^{\frac{π}{2}} \frac{sinx}{1 + \cos^{2} x} dx$

Ans

$\begin{array}{l} Let I = \int_{0}^{\frac{π}{2}} \frac{sinx}{1 + \cos^{2} x} dx \\ Again, let t = cosx \Rightarrow \frac{dt}{dx} = - sin x \\ When x = 0, t = 1 and when x = \frac{π}{2}, t = 0 \\ ∴ I = \int_{1}^{0} \frac{sinx}{1 + t^{2}} \frac{dt}{- sinx} \\ = - \int_{1}^{0} \frac{1}{1 + t^{2}} dt \\ = - {[\tan^{- 1} t]}_{1}^{0} \\ = - (\tan^{- 1} 0 - \tan^{- 1} 1) \\ = - (0 - \frac{π}{4}) = \frac{π}{4} \end{array}$

Q.5

$Evaluate the integrals \int_{0}^{2} \frac{dx}{x + 4 - x^{2}}$

Ans

$\begin{array}{l} Let I = \int_{0}^{2} \frac{dx}{x + 4 - x^{2}} \\ = \int_{0}^{2} \frac{dx}{- (x^{2} - x - 4)} \\ = \int_{0}^{2} \frac{dx}{- (x^{2} - x + \frac{1}{4} - \frac{1}{4} - 4)} \\ = - \int_{0}^{2} \frac{dx}{{(x - \frac{1}{2})}^{2} - \frac{17}{4}} \\ = - \int_{0}^{2} \frac{dx}{{(x - \frac{1}{2})}^{2} - {(\frac{\sqrt{17}}{4})}^{2}} \\ Let t = x - \frac{1}{2} \Rightarrow \frac{dt}{dx} = 1 \\ When x = 0, t = - \frac{1}{2} and when x = 2, t = \frac{3}{2} \\ I = - \int_{- \frac{1}{2}}^{\frac{3}{2}} \frac{dt}{{(\frac{\sqrt{17}}{4})}^{2} - t^{2}} \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} \log {\frac{(\frac{\sqrt{17}}{4}) + t}{(\frac{\sqrt{17}}{4}) - t}}_{- \frac{1}{2}}^{\frac{3}{2}} \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} [\log {\frac{(\frac{\sqrt{17}}{4}) + \frac{3}{2}}{(\frac{\sqrt{17}}{4}) - \frac{3}{2}}} - \log {\frac{(\frac{\sqrt{17}}{4}) - \frac{1}{2}}{(\frac{\sqrt{17}}{4}) + \frac{1}{2}}}] \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} [\log {\frac{\sqrt{17} + 3}{\sqrt{17} - 3}} - \log {\frac{\sqrt{17} - 1}{\sqrt{17} + 1}}] \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} \log {\frac{\sqrt{17} + 3}{\sqrt{17} - 3} \times \frac{\sqrt{17} + 1}{\sqrt{17} - 1}} \\ = \frac{1}{2 (\frac{\sqrt{17}}{4})} \log {\frac{17 + \sqrt{17} + 3 \sqrt{17} + 3}{17 - \sqrt{17} - 3 \sqrt{17} + 3}} \\ = \frac{2}{\sqrt{17}} \log {\frac{20 + 4 \sqrt{17}}{20 - 4 \sqrt{17}}} \\ = \frac{2}{\sqrt{17}} \log {\frac{5 + \sqrt{17}}{5 - \sqrt{17}}} \\ = \frac{2}{\sqrt{17}} \log {\frac{5 + \sqrt{17}}{5 - \sqrt{17}} \times \frac{5 + \sqrt{17}}{5 + \sqrt{17}}} \\ = \frac{2}{\sqrt{17}} \log {\frac{25 + 17 + 10 \sqrt{17}}{25 - 17}} \\ = \frac{2}{\sqrt{17}} \log {\frac{42 + 10 \sqrt{17}}{8}} \\ = \frac{2}{\sqrt{17}} \log {\frac{21 + 5 \sqrt{17}}{4}} \end{array}$

Q.6

$Evaluate the integrals \int_{- 1}^{1} \frac{dx}{x^{2} + 2 x + 5} dx$

Ans

$\begin{array}{l} \int_{- 1}^{1} \frac{dx}{x^{2} + 2 x + 5} dx = \int_{- 1}^{1} \frac{dx}{{(x + 1)}^{2} + 2^{2}} dx \\ Let t = x + 1 \Rightarrow \frac{dt}{dx} = 1 \\ When x = - 1, t = 0 and when x = 1, t = 2 \\ ∴ \int_{- 1}^{1} \frac{dx}{{(x + 1)}^{2} + 2^{2}} dx = \int_{0}^{2} \frac{dx}{t^{2} + 2^{2}} dt \\ = {\frac{1}{2} \tan^{- 1} (\frac{t}{2})}_{0}^{2} \\ = \frac{1}{2} \tan^{- 1} (\frac{2}{2}) - \frac{1}{2} \tan^{- 1} (\frac{0}{2}) \\ = \frac{1}{2} \tan^{- 1} (1) - \frac{1}{2} \tan^{- 1} (0) \\ = \frac{1}{2} \times \frac{π}{4} - \frac{1}{2} \times 0 = \frac{π}{8} \\ Thus, \int_{- 1}^{1} \frac{dx}{x^{2} + 2 x + 5} dx = \frac{π}{8} \end{array}$

Q.7

$Evaluate the integrals \int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x^{2}}) e^{2 x} dx$

Ans

$\begin{array}{l} Let I = \int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x^{2}}) e^{2 x} dx \\ = \int_{1}^{2} (\frac{2}{2 x} - \frac{2}{{(2 x)}^{2}}) e^{2 x} dx \\ Again let t = 2 x \Rightarrow \frac{dt}{dx} = 2 \\ When x = 1, t = 2 and when x = 2, t = 4 \\ I = \int_{2}^{4} (\frac{2}{t} - \frac{2}{t^{2}}) e^{t} \frac{dt}{2} \\ = \int_{2}^{4} (\frac{1}{t} - \frac{1}{t^{2}}) e^{t} dt \\ Let F (t) = \frac{1}{t}, Then F ‘ (t) = - \frac{1}{t^{2}} \\ \Rightarrow \int_{2}^{4} (\frac{1}{t} - \frac{1}{t^{2}}) e^{t} dt = \int_{2}^{4} (F (t) - F ‘ (t)) e^{t} dt \\ = {[e^{t} F (t)]}_{2}^{4} \\ = {[e^{t} . \frac{1}{t}]}_{2}^{4} \\ = e^{4} . \frac{2}{4} - e^{2} . \frac{1}{2} \\ = \frac{1}{4} e^{2} (e^{2} - 2) \\ Thus, \\ \int_{1}^{2} (\frac{1}{x} - \frac{1}{2 x^{2}}) e^{2 x} dx = \frac{1}{4} e^{2} (e^{2} - 2) \end{array}$

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To leave a chapter on the CBSE Board for Class 12 is extremely risky. Chapter 7 is on integrals, and as seen in the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, these concepts are extremely essential for students to prepare. Teachers note that the primary reason for this is that the fundamental concepts of integration and differentiations are not well understood on the student’s end. Therefore, for a student to understand and work on this query, they need to strictly follow the NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10. The explanations that are provided explain each step in the most elemental and simplified way. The more a student practises the exercises and goes through all the explanations in NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, the more they will slowly regain their confidence and gain a good comprehension of the chapter. The NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.10, were made available to Extramarks to address issues like this among students; thus, skipping the revision of the chapter is highly unadvisable.

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.10) Exercise 7.10

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.10) Exercise 7.10

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NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.10) Exercise 7.10

NCERT Solutions for Class 12 Maths Chapter 7 Integrals (Ex 7.10) Exercise 7.10

Access NCERT Solutions for Class 12 Maths Chapter 7 – Integrals

What does the NCERT Class 12 Chapter 7 Exercise 7.10 Contain?

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