NCERT Solutions Class 12 maths Chapter-7 (Integrals)Exercise 7.7
NCERT Solutions Class 12 Maths from class
12th Students will get the answers of
Chapter-7 (Integrals)Exercise 7.7 This chapter will help you to learn the basics and you should expect at least one question in your exam from this chapter.
We have given the answers of all the questions of
NCERT Board Mathematics Textbook in very easy language, which will be very easy for the students to understand and remember so that you can pass with good marks in your examination.
Exercise 7.7
Q1. Integrate the function :√4−x2
Answer. Let I=∫√4−x2dx=∫√(2)2−(x)2dx It is known that, ∫√a2−x2dx=x2√a2−x2a22sin−1xa+C∴I=x2√4−x2+42sin−1x2+C=x2√4−x2+2sin−1x2+C
Q2. Integrate the function : √1−4x2
Answer. Let I=∫√1−4x2dx=∫√(1)2−(2x)2dx Let 2x=t⇒2dx=dt∴I=12∫√(1)2−(t)2dt It is known that, ∫√a2−x2dx=x2√a2−x2+a22sin−1xa+C⇒I=12[t2√1−t2+12sin−1t]+C =t4√1−t2+14sin−1t+C=2x4√1−4x2+14sin−12x+C=x2√1−4x2+14sin−12x+C
Q3. Integrate the function : √x2+4x+6
Answer. Let I=∫√x2+4x+6dx=∫√x2+4x+4+2dx=∫√(x2+4x+4)+2dx=∫√(x+2)2+(√2)2dx It is known that,∫√x2+a2dx=x2√x2+a2+a22log∣∣x+√x2+a2∣∣+C ∴I=(x+2)2√x2+4x+6+22log(x+2)+√x2+4x+6|+C=(x+2)2√x2+4x+6+log(x+2)+√x2+4x+6|+C
Q4. Integrate the function :√x2+4x+1x2+4x+1
Answer. Let I=∫√x2+4x+1dx=∫√(x2+4x+4)−3dx=∫√(x+2)2−(√3)2dx It is known that,∫√x2−a2dx=x2√x2−a2−a22log∣∣x+√x2−a2∣∣+C ∴I=(x+2)2√x2+4x+1−32log∣∣(x+2)+√x2+4x+1∣∣+C
Q5. Integrate the function : √1−4x−x2
Answer. Let I=∫√1−4x−x2dx=∫√1−(x2+4x+4−4)dx=∫√1+4−(x+2)2dx=∫√(√5)2−(x+2)2dx It is a known that,∫√a2−x2dx=x2√a2−x2+a22sin−1xa+C ∴I=(x+2)2√x2+4x−5−92log(x+2)+√x2+4x−5|+C
Q6. Integrate the function : √x2+4x−5
Answer. Let I=∫√x2+4x−5dx=∫√(x2+4x+4)−9dx=∫√(x+2)2−(3)2dx It is known that, ∫√x2−a2dx=x2√x2−a2−a22log∣∣x+√x2−a2∣∣+C∴I=(x+2)2√x2+4x−5−92log(x+2)+√x2+4x−5|+C
Q7. Integrate the function :√1+3x−x2
Answer. Let I=∫√1+3x−x2dx=∫√1−(x2−3x+94−94)dx=∫√(1+94)−(x−32)2dx=∫
⎷(√132)2−(x−32)2dx It is known that, ∫√a2−x2dx=x2√a2−x2+a22sin−1xa+C∴I=32√1+3x−x2+134×2sin−1(x−32√132)+C=2x−34√1+3x−x2+138sin−1(2x−3√13)+C
Q8. Integrate the function :8.√x2+3x
Answer. Let I=∫√x2+3xdx=∫√x2+3x+94−94dx=∫√(x+32)2−(32)2dx It is known that, ∫√x2−a2dx=x2√x2−a2−a22log∣∣x+√x2−a2∣∣+C∴I=(x+32)2√x2+3x−942log(x+32)+√x2+3x|+C=(2x+3)4√x2+3x−98log(x+32)+√x2+3x|+C
Q9. Integrate the function : √1+x29
Answer. Let I=∫√1+x29dx=13∫√9+x2dx=13∫√(3)2+x2dx It is known that, ∫√x2+a2dx=x2√x2+a2+a22log∣∣x+√x2+a2∣∣+C∴I=13[x2√x2+9+92log∣∣x+√x2+9∣∣]+C=x6√x2+9+32log∣∣x+√x2+9∣∣+C
Q10. Choose the correct answer : ∫√1+x2 equals to (A) x2√1+x2+12log(x+√1+x2)|+C (B) 23(1+x2)32+C (C) 23x(1+x2)32+C (D)x22√1+x2+12x2log∣∣x+√1+x2∣∣+C(D)x221+x2+12x2log|x+1+x2|+C
Answer. It is known that, ∫√a2+x2dx=x2√a2+x2+a22log∣∣x+√x2+a2∣∣+C∴∫√1+x2dx=x2√1+x2+12log∣∣x+√1+x2∣∣+C Hence, the correct Answer is A .
Q11. Choose the correct answer : ∫√x2−8x+7 equal to (A) 12(x−4)√x2−8x+7+9log∣∣x−4+√x2−8x+7∣∣+C (B) 12(x+4)√x2−8x+7+9log∣∣x+4+√x2−8x+7∣∣+C (C) 12(x−4)√x2−8x+7−3√2log∣∣x−4+√x2−8x+7∣∣+C (D) 12(x−4)√x2−8x+7−92log∣∣x−4+√x2−8x+7∣∣+C
Answer. Let I=∫√x2−8x+7dx=∫√(x2−8x+16)−9dx=∫√(x−4)2−(3)2dx It is known that, ∫√x2−a2dx=x2√x2−a2−a22log∣∣x+√x2−a2∣∣+C ∴I=(x−4)2√x2−8x+7−92log∣∣(x−4)+√x2−8x+7∣∣+C Hence, the correct Answer is D.
Chapter-7 (integrals)
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